% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
%                                                                         %
% The Project Gutenberg EBook of Short Cuts in Figures, by A. Frederick Collins
%                                                                         %
% This eBook is for the use of anyone anywhere at no cost and with        %
% almost no restrictions whatsoever.  You may copy it, give it away or    %
% re-use it under the terms of the Project Gutenberg License included     %
% with this eBook or online at www.gutenberg.org                          %
%                                                                         %
%                                                                         %
% Title: Short Cuts in Figures                                            %
%        to which is added many useful tables and formulas written        %
%        so that he who runs may read                                     %
%                                                                         %
% Author: A. Frederick Collins                                            %
%                                                                         %
% Release Date: September 6, 2009 [EBook #29914]                          %
%                                                                         %
% Language: English                                                       %
%                                                                         %
% Character set encoding: ISO-8859-1                                      %
%                                                                         %
% *** START OF THIS PROJECT GUTENBERG EBOOK SHORT CUTS IN FIGURES ***     %
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\begin{verbatim}
The Project Gutenberg EBook of Short Cuts in Figures, by A. Frederick Collins

This eBook is for the use of anyone anywhere at no cost and with
almost no restrictions whatsoever.  You may copy it, give it away or
re-use it under the terms of the Project Gutenberg License included
with this eBook or online at www.gutenberg.org


Title: Short Cuts in Figures
       to which is added many useful tables and formulas written
       so that he who runs may read

Author: A. Frederick Collins

Release Date: September 6, 2009 [EBook #29914]

Language: English

Character set encoding: ISO-8859-1

*** START OF THIS PROJECT GUTENBERG EBOOK SHORT CUTS IN FIGURES ***
\end{verbatim}

\vfill

\frontmatter

\begin{center}
\bfseries

\nbvspace[1]
\TitleHuge
{\nbtitlestretch
SHORT CUTS IN\\
FIGURES}

\nbvspace[1]
\normalsize
TO WHICH IS ADDED MANY USEFUL TABLES\\
AND FORMULAS WRITTEN SO THAT\\
HE WHO RUNS MAY READ

\nbvspace[1]
\small BY\\
\Large A. FREDERICK COLLINS \\[0.5em]
\footnotesize AUTHOR OF ``A WORKING ALGEBRA,'' ``WIRELESS TELEGRAPHY,\\
ITS HISTORY, THEORY AND PRACTICE,'' ETC., ETC.

\nbvspace[2]
\includegraphics[width=0.8in]{./images/ejc.pdf}

\nbvspace[3]
\normalsize
NEW YORK

\Large
EDWARD J. CLODE
\nbvspace[1]

\end{center}
\clearpage
%-----File: 002.png-----

\begin{center}

\nbvspace[2]
COPYRIGHT, 1916, BY\\
EDWARD J. CLODE

\nbvspace[3]
\footnotesize
\begin{tabular}{@{}c@{}}
\hline\\[-1ex]
PRINTED IN THE UNITED STATES OF AMERICA
\end{tabular}

\end{center}
\clearpage
%-----File: 003.png-----

\begin{center}

\nbvspace[2]
\footnotesize
TO\\[0.5ex]
\small
WILLIAM H. BANDY\\[1ex]
\footnotesize
AN EXPERT AT SHORT CUTS\\
IN FIGURES\\

\nbvspace[3]
\end{center}
\clearpage
\thispagestyle{empty}
\begin{center}
\small
Produced by Peter Vachuska, Nigel Blower and the Online
Distributed Proofreading Team at http://www.pgdp.net
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  Detailed \hyperref[section:tn]{\textit{Transcriber's Notes}} may be found at the end of this document.
}

\clearpage
%%%%%%%%% End of Production Credits and Transcriber's Notes %%%%%%%%%
%-----File: 004.png-----

\pdfbookmark[0]{A Word to You}{A Word to You}
\chapter*{A WORD TO YOU}

\textsc{Figuring} is the key-note of all business. To
know how to figure quickly and accurately is to
jack-up the power of your mind, and hence your
efficiency, and the purpose of this book is to tell
you how to do it.

Any one who can do ordinary arithmetic can
easily master the simple methods I have given to
figure the right way as well as to use short cuts,
and these when taken together are great savers
of time and effort and, consequently, of money.

Not to be able to work examples by the most
approved short-cut methods known to mathematical
science is a tremendous handicap and if you
are carrying this kind of a dead weight get rid
of it at once or you will be held back in your race
for the grand prize of success.

On the other hand, if you are quick and accurate
at figures you wield a tool of mighty power and
importance in the business world, and then by
making use of short cuts you put a razor edge on
that tool with the result that it will cut fast and
smooth and sure, and this gives you power multiplied.

Should you happen to be one of the great majority
who find figuring a hard and tedious task it is
%-----File: 005.png-----
simply because you were wrongly taught, or taught
not at all, the fundamental principles of calculation.

By following the simple instructions herein
given you can correct this fault and not only
learn the true methods of performing ordinary
operations in arithmetic but also the proper use
of scientific short cuts by means of which you can
achieve both speed and certainty in your work.

Here then, you have a key which will unlock
the door to rapid calculation and all you have to
do, whatever vocation you may be engaged in,
is to enter and use it with pleasure and profit.

\begin{flushright}
\small A. FREDERICK COLLINS
\end{flushright}
\cleardoublepage
%-----File: 006.png-----

\pdfbookmark[0]{Contents}{Contents}
\tableofcontents

\cleardoublepage
%-----File: 007.png-----

\begin{center}
\nbvspace[2]
\Large\bfseries SHORT CUTS IN FIGURES
\nbvspace[3]
\end{center}
\cleardoublepage
%-----File: 008.png-----

\mainmatter

\pagestyle{fancy}

\nbchapter{I}{What Arithmetic Is}
\index{Arithmetic, definition of}
\index{Definition of arithmetic}

\begin{nbchaptercontentstwo}
The Origin of Calculation.&Ratios and Proportions.\\
The Origin of Counting and Figures.&Practical Applications of Arithmetic.\\
Other Signs Used in Arithmetic.&  \quad Percentage.\\
The Four Ground Rules.&  \quad Interest.\\
The Operation of Addition.&  \quad \quad Simple Interest.\\
The Operation of Multiplication.&Compound Interest.\\
The Operation of Division.&  \quad Profit and Loss.\\
The Operation of Subtraction.&  \quad \quad Gross Profit.\\
Fractions.&  \quad \quad Net Profit.\\
Decimals.&  \quad \quad Loss.\\
Powers and Roots.&Reduction of Weights and Measures.
\end{nbchaptercontentstwo}

\nbparagraph{The Origin of Calculation.}%
\index{Origin of calculation}---To be able to
figure in the easiest way and in the shortest time
you should have a clear idea of what arithmetic is
and of the ordinary methods used in calculation.

To begin with arithmetic means that we take
certain numbers we already know about, that is
the \textit{value} of, and by manipulating them, that is
performing an operation with them, we are able to
find some number which we do not know but
which we want to know.

Now our ideas about numbers are based entirely
on our ability to measure things and this in turn
is founded on the needs of our daily lives.
%-----File: 009.png-----

To make these statements clear suppose that
\textit{distance} did not concern us and that it would not
take a longer \textit{time} or greater \textit{effort} to walk a mile
than it would to walk a block. If such a state of
affairs had always existed then primitive man\index{Primitive man}
never would have needed to judge that a day's
walk was once again as far as half a day's walk.

In his simple reckonings he performed not only
\index{Origin of addition}\index{Addition, operation of|(}\index{Operation of addition|(}%
\textit{the operation of addition} but he also laid the foundation
for the measurement of \textit{time}.

Likewise when primitive man considered the
\index{Origin of subtraction}\index{Subtraction, operation of|(}\index{Operation of subtraction|(}%
difference in the length of two paths which led,
let us say, from his cave to the pool where the
mastodons came to drink, and he gauged them so
that he could choose the shortest way, he performed
the operation of \textit{subtraction} though he did not
work it out arithmetically, for figures had yet to
be invented.

And so it was with his food. The scarcity of it\index{Origin of fractions}
made the Stone Age man lay in a supply to tide
over his wants until he could replenish his stock;
and if he had a family he meted out an equal portion
of each delicacy to each member, and in this
way the fractional measurement of things came
about.

There are three general divisions of measurements\index{Measurement, divisions of}\index{Measurement of time, length, and mass}
and these are (1)~the measurement of
\textit{time}; (2)~the measurement of \textit{space} and (3)~the
measurement of \textit{matter}; and on these three fundamental
elements of nature through which all
%-----File: 010.png-----
phenomena are manifested to us arithmetical operations
of every kind are based if the calculations
are of any practical use.\footnote
  {To measure \textit{time}, \textit{space}, and \textit{matter}, or as these elements are called
  in physics \textit{time}, \textit{length}, and \textit{mass}, each must have a unit of its own so
  that other quantities of a like kind can be compared with them. Thus
  the unit of time is the \textit{second}; the unit of length is the \textit{foot}, and the unit
  of mass is the \textit{pound}, %[**TN: pound italicised]
  hence these form what is called the \textit{foot-pound-second}\index{Foot-pound-second system}
  system. All other units relating to motion and force may be
  easily obtained from the F.P.S. system.}

\nbparagraph{The Origin of Counting and Figures.}%
\index{Counting, origin of}\index{Origin of counting}\index{Figures, origin of}\index{Origin of figures}%
---As civilization
grew on apace it was not enough for man
to measure things by comparing them roughly with
other things which formed his units, by the sense
of sight or the physical efforts involved, in order
to accomplish a certain result, as did his savage
forefathers.

And so \textit{counting}, or \textit{enumeration}\index{Enumeration, origin of} as it is called,
was invented, and since man had five digits\footnote
  {The word \textit{digit}\index{Digits, meaning of} means any one of the terminal members of the
  hand including the thumb, whereas the word \textit{finger} excludes the thumb.
  Each of the Arabic numerals\index{Arabic numerals}, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $0$, is called a \textit{digit}
  and is so named in virtue of the fact that the fingers were first used to
  count upon.}
on each hand it was the most natural thing in the
world that he should have learned to count on his
digits, and children still very often use their digits
for this purpose and occasionally grown-ups too.

Having made each digit a unit, or \textit{integer}\index{Integers} as
it is called, the next step was to give each one a
definite name to call the unit by, and then came
the writing of each one, not in unwieldy words
%-----File: 011.png-----
but by a simple mark, or a combination of marks
called a \textit{sign} or \textit{symbol}, and which as it has come
down to us is $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $0$.

By the time man had progressed far enough to
name and write the symbols for the units he had
two of the four \textit{ground rules}\index{Ground rules}, or
\textit{fundamental operations}\index{Fundamental operations}
as they are called, well in mind, as well as the
combination of two or more figures to form numbers
as $10$, $23$, $108$, etc.

\nbparagraph{Other Signs Used in Arithmetic.}%
\index{Arithmetical signs and symbols}\index{Signs, arithmetical}\index{Symbols, arithmetical}%
---Besides the
symbols used to denote the figures there are symbols
employed to show what arithmetical operation
is to be performed.

$+\;$ Called \textit{plus}\index{Plus sign}. It is the sign of \textit{addition}; that is,
it shows that two or more figures or numbers are
to be added to make more, or to find the sum of
them, as $5 + 10$. The plus sign was invented by
Michael Stipel in 1544 and was used by him in
his \textit{Arithmetica Integra}.

$=\;$ Called \textit{equal}\index{Equal sign}. It is the sign of \textit{equality} and
it shows that the numbers on each side of it
are of the same amount or are of \textit{equal value}, as
$5 + 10 = 15$. The sign of equality was published
for the first time by Robert Recorde in 1557, who
used it in his algebra.

$-\;$ Called \textit{minus}\index{Minus sign}. It is the sign of \textit{subtraction} and
it shows that a number is to be taken away or
\textit{subtracted} from another given number, as $10-5$.
The minus sign was also invented by Michael
Stipel.
%-----File: 012.png-----

$×\;$ Called \textit{times}\index{Multiplication sign}\index{Times sign}.
It is the sign of \textit{multiplication}
and means \textit{multiplied by}; that is, taking one number
as many times as there are units in the other,
thus, $5 × 10$. The sign of multiplication was devised
by William Oughtred in 1631. It was
called St.~Andrew's Cross and was first published
in a work called \textit{Clavis Mathematicae}, or Key to
Mathematics.

$÷\;$ Called the \textit{division sign}\index{Division sign}. It is the sign of
division and means \textit{divided by}; that is, it shows a
given number is to be contained in, or divided by,
another given number, as $10 ÷ 5$. The division
sign was originated by Dr.~John Pell, a professor
of mathematics and philosophy.

\nbparagraph{The Four Ground Rules.}\index{Ground rules}%
---In arithmetic the
operations of addition, subtraction, multiplication,
and division are called the \textit{ground rules} because
all other operations such as fractions, extracting
roots, etc., are worked out by them.

\nbparagraph{The Operation of Addition.}%
\index{Addition, operation of|)}\index{Operation of addition|)}
---The ordinary definition
of addition is the operation of finding a number
which is equal to the value of two or more
numbers.

This means that in addition we start with an
unknown quantity which is made up of two or
more known parts and by operating on these parts
in a certain way we are able to find out exactly
what the whole number of parts, or the unknown
quantity, is.

To simply count up the numbers of parts is not
%-----File: 013.png-----
enough to perform the operation of addition, for
when this is done we still have an unknown quantity.
But to actually find what the \textit{whole} number\index{Whole number}
of parts is, or the \textit{sum} of them as it is called, we
have to count off all of the units of all of the numbers,
thus:
\[
2 + 2 + 2 + 2 = 8
\]

Now when arithmetic began prehistoric man\index{Prehistoric man}
had to add these parts by using his fingers and
saying $1$ and $1$ are $2$, and $1$ are $3$, and $1$ are $4$, and
so on, adding up each unit until he got the desired
result.

Some time after, and it was probably a good
many thousands of years, an improvement was
made in figuring and the operation was shortened
so that it was only necessary to say $2 + 2$ are $4$,
and $2$ are $6$, and $2$ are $8$. When man was able to
add four $2$'s without having to count each unit
in each number he had made wonderful progress
and it could not have taken him long to learn to
add up other figures in the same way.

\nbparagraph{The Operation of Subtraction.}%
\index{Subtraction, operation of|)}\index{Operation of subtraction|)}%
---Subtraction is
the operation of taking one number from the other
and finding the difference between them.

To define subtraction in another and more
simple way, we can say that it is the operation of
starting with a known quantity which is made up
of two or more parts and by taking a given number
of parts from it we can find what the difference or
remainder is in known parts. Hence the operation
%-----File: 014.png-----
of subtraction is just the inverse of that of addition.

The processes of the mind which lead up to the
operation of subtraction are these: when man
began to concern himself with figures and he
wanted to take $4$ away from $8$ and to still know
how many remained he had to count the units
that were left thus:
\[
1 + 1 + 1 + 1 = 4
\]

And finally, when he was able through a better
understanding of figures and with a deal of practice
to say $8 - 4 = 4$, without having to work it
all out in units, he had made a great stride and
laid down the second ground rule of arithmetic.
But, curiously enough, the best method in use
at the present time for performing the operation
of subtraction is by addition, which is a reversion
to first principles, as we shall presently see in
\hyperref[chapter:III]{Chapter~III}.

\nbparagraph{The Operation of Multiplication.}%
\index{Multiplication, operation of}\index{Operation of multiplication}%
---The rule for
\textit{multiplication} states that it is the operation of
taking one number as many times as there are
units in another.

In the beginning of arithmetic, when one number
was to be taken as many times as there were
units in another, the \textit{product} was obtained by
cumulative addition\index{Addition, cumulative}\index{Cumulative addition},
the figures being added together
thus:
\[
7 + 7 + 7 + 7 + 7 = 35
\]
%-----File: 015.png-----

Having once found that $7$ taken $5$ times gave
the product $35$ it was a far easier mental process
to remember the fact, namely, that
\[
7 × 5 = 35
\]
than it was to add up the five $7$'s each time; that
is if the operation had to be done very often, and
so another great short cut was made in the operation
of addition and mental calculation took
another step forward.

But multiplication was not only a mere matter
of memorizing the fact that $7 × 5 = 35$ but it meant
that at least $100$ other like operations had to be
remembered and this resulted in the invention of
that very useful arithmetical aid---the \textit{multiplication
table}\index{Multiplication table}.

While multiplication is a decided short cut in
solving certain problems in addition, it is a great
deal more than addition for it makes use of the
relation, or \textit{ratio}\index{Ratio} as it is called, between two numbers
or two quantities of the same kind and this
enables complex problems to be performed in an
easy and rapid manner.

Hence the necessity for the absolute mastery of
the multiplication table, as this is the master-key
which unlocks many of the hardest arithmetical
problems.

A quick memory and the multiplication table
well learned will bring about a result so that the
product of any two factors will be on the tip of
%-----File: 016.png-----
your tongue or at the point of your pencil, and
this will insure a rapidity of calculation that cannot
be had in any other way.

\nbparagraph{The Operation of Division.}%
\index{Division, operation of}\index{Operation of division}%
---Division is the
operation of finding one of two numbers called the
\textit{factors}\index{Factors}, that is, the \textit{divisor}\index{Divisor} and the \textit{quotient}\index{Quotient}, when
the whole number, or \textit{dividend}\index{Dividend} as it is called, and
the factor called the divisor are known.

Defined in more simple terms, division is the
operation of finding how many times one number
is contained in another number. Hence division,
it will be seen, is simply the inverse operation of
multiplication.

Since division and multiplication are so closely
related it would seem that division should not
have been very hard to learn in the beginning but
\textit{long division}\index{Division, long} was, nevertheless, an operation that
could only be done by an expert arithmetician.

It will make division an easier operation if it is
kept in mind that it is the inverse of multiplication;
that is, the operation of division annuls the operation
of multiplication, since if we multiply $4$ by $3$
we get $12$ and if we divide $12$ by $3$ we get $4$, and
we are back to the place we started from. For this
reason problems in division can be proved by
multiplication and conversely multiplication can
be proved by division.

\nbparagraph{Fractions.}\index{Fractions}%
---A fraction is any part of a whole
number or unit. While a whole number may be
divided into any number of fractional parts the
%-----File: 017.png-----
fractions are in themselves numbers just the same.
Without fractions there could be no measurement
and the more numerous the fractional divisions of
a thing the more accurately it can be measured.

From the moment that man began to measure
off distances and quantities he began to use fractions,
and so if fractions were not used before whole
numbers they were certainly used concurrently with
them. In fact the idea of a whole number is made
clearer to the mind by thinking of a number of
parts as making up the whole than by considering
the whole as a unit in itself.

It will be seen then that while fractions are
parts of whole numbers they are in themselves
numbers and as such they are subject to the same
treatment as whole numbers; that is operations
based on the four ground rules, namely, addition,
subtraction, multiplication, and division.

Fractions may be divided into two general
classes and these are (1)~\textit{common}\index{Common fractions}\index{Fractions, common}
or \textit{vulgar}\footnote
  {Once upon a time anything that was common or ordinary was
  called \textit{vulgar}, hence common fractions were and are still called vulgar
  fractions.}
\textit{fractions}\index{Vulgar fractions}\index{Fractions, vulgar}
and (2)~\textit{decimal fractions}\index{Decimal fractions}\index{Fractions, decimal}. Vulgar fractions
may be further divided into (A)~\textit{proper fractions}\index{Proper fractions}\index{Fractions, proper}
and (B)~\textit{improper fractions}\index{Improper fractions}\index{Fractions, improper},
and both vulgar and
decimal fractions may be operated as (a)~\textit{simple
fractions}\index{Simple fractions}\index{Fractions, simple},
(b)~\textit{compound fractions}\index{Compound fractions},
and (c)~\textit{complex
fractions}\index{Complex fractions}\index{Fractions, complex},
the latter including \textit{continued fractions}\index{Continued fractions}\index{Fractions, continued},
all of which is explained in \hyperref[chapter:VI]{Chapter~VI}.
%-----File: 018.png-----

\nbparagraph{Decimals.}\index{Decimals}---Since there are five digits on each
hand it is easy to see how the \textit{decimal system}\index{Decimal system}, in
which numbers are grouped into tens, had its
origin.

The word \textit{decimal} means $10$ and decimal arithmetic
is based on the number $10$; that is, all operations
use powers of $10$ or of $\frac{1}{10}$. But instead of
writing the terms down in vulgar fractions as $\frac{1}{10}$,
$\frac{5}{10}$, or $\frac{5}{100}$, these terms are expressed as whole
numbers thus $.1$, $.5$, $.05$ when the fractional value
is made known by the position of the \textit{decimal point}\index{Decimal point}.

This being true the four fundamental operations
may be proceeded with just as though whole numbers
were being used and this, of course, greatly
simplifies all calculations where fractions are factors,
that is, provided the decimal system can be used at
all.

It is not often, though, except in calculations
involving money or where the \textit{metric system}\index{Metric system}\footnote
  {The metric system of weights and measures is described and tables
  are given in \hyperref[chapter:VIII]{Chapter~VIII}.}
of weights and measures is used, that the decimal
system can be applied with exactness, for few
common fractions can be stated exactly by them;
that is, few common fractions can be changed to
decimal fractions and not leave a remainder.

\nbparagraph{Powers and Roots.}%
\index{Powers and roots}%
---\nbsubparagraph{Powers}---By \textit{involution}\index{Involution}, or
\textit{powers}\index{Powers}, is meant an operation in which a number is
multiplied by itself, as $2 × 2 = 4$, $10 × 10 = 100$,
etc.
%-----File: 019.png-----

The number to be multiplied by itself is called
the \textit{number}\index{Number in powers}; the number by which it is multiplied
is called a \textit{factor}\index{Factor in Powers}, and the number obtained by
multiplying is called the \textit{power}, thus:
\begin{align*}
&\phantom{\text{nu}\nearrow 2 × 2}\raisebox{-0.5ex}{$\swarrow$\;} \text{factor}\\[-0.5ex]
&\phantom{\text{nu}\nearrow} 2 × 2 = 4 \;\leftarrow\; \text{square, or second power.}\\[-1ex]
&\phantom{\text{nu}}\raisebox{0.3ex}{$\nearrow$}\\[-1ex]
&\text{number}
\end{align*}
In algebra\index{Algebra defined}, which is a kind of generalized shorthand
arithmetic, it is written
\begin{align*}
&\phantom{\text{nu}\nearrow 2}\raisebox{-0.2ex}{\;$\swarrow$\;} \text{index}\\[-0.5ex]
&\phantom{\text{nu}\nearrow} 2^2 = 4 \;\leftarrow\; \text{square, or second power.}\\[-1ex]
&\phantom{\text{nu}}\raisebox{0.3ex}{$\nearrow$}\\[-1ex]
&\text{number}
\end{align*}
where $2$ is the number, $^2$ is the factor, called the
\textit{index}\index{Index in powers}, and $4$ is the square, or second power.

\nbsubparagraph{Roots}.\index{Roots}---By \textit{evolution}\index{Evolution},
or the \textit{extraction of roots}\index{Extraction of roots}
as it is called, is meant the operation of division
whereby a number divided by another number
called the factor will give another number called
its root, as
\begin{align*}
&\phantom{\text{nu}\nearrow 16 ÷ 4}\raisebox{-0.5ex}{$\swarrow$\;} \text{factor}\\[-0.5ex]
&\phantom{\text{nu}\nearrow} 16 ÷ 4 = 4 \;\leftarrow\; \text{square root}\\[-1ex]
&\phantom{\text{nu}}\raisebox{0.3ex}{$\nearrow$}\\[-1ex]
&\text{number}
\end{align*}

In algebra it is written
\begin{align*}
&\phantom{\text{radical sign }\nearrow}\sqrt{16} = 4 \;\leftarrow\; \text{square root}\\[-0.5ex]
&\text{radical sign\index{Radical sign} }\raisebox{0.5ex}{$\nearrow$} \qquad \raisebox{0.5ex}{$\nwarrow$} \text{ number}
\end{align*}

Hence the extraction of roots is the inverse of
%-----File: 020.png-----
multiplication but it is a much more difficult
operation to perform than that of involution.

\nbparagraph{Ratios and Proportions.}---\nbsubparagraph{Ratio}\index{Ratio}
is the relation
which one number or quantity bears to another
number or quantity of the same kind, as $2$ to $4$,
$3$ to $5$, etc.

To find the ratio is simply a matter of division,
that is, one number or quantity is divided by
another number or quantity, and the resulting
quotient\index{Quotient in ratios} is the proper relation between the two
numbers or quantities.

\nbsubparagraph{Proportion}\index{Proportion} is the equality between two ratios or
quotients as, $2$ is to $4$ as $3$ is to $6$, or $3$ is to $4$ as
$75$ is to $100$, when the former is written
\begin{flalign*}
&&2:4 &:: 3:6\\
\text{and the latter}&&3:4 &:: 75:100&&\phantom{\text{and the latter}}
\end{flalign*}

In the last named case $3:4$ can be written $\frac{3}{4}$ and
$75: 100$ can be written $\frac{75}{100}$. Then $\frac{3}{4} = \frac{75}{100}$ and this
equality between the two numbers may be proved
to be true because $\frac{75}{100}$ may be reduced to $\frac{3}{4}$.

The operation of proportion is largely used in
business calculations. As an illustration suppose
$4$ yards of silk cost \$1.00 and you want to find the
price of $3$ yards at the same rate, then
\begin{flalign*}
&&3\text{ yds.}: 4\text{ yds.} &:: \text{ \$?} : \text{ \$}1.00\;\footnotemark \\
\text{or}&&       \tfrac{3}{4} &= \tfrac{?}{100}&&\phantom{\text{or}}
\end{flalign*}%
  \footnotetext{In the above example in proportion a question mark is used as the
  symbol for the unknown term which we wish to find. In algebra $x$\index{X in algebra} %[**TN: original wording "X in albegra"]
  is generally used to denote unknown quantities and is a more correct
  mode of expression.}%
%-----File: 021.png-----
now cross-multiplying\index{Cross-multiplying}\index{Multiplying, Cross-} we have
\[
300 = 4?
\]
or the cost of 3 yards is equal to
\[
3.00 ÷ 4\text{ or }75\text{ cents.}
\]

{\nbparagraph{Practical Applications.}---The every-day applications
of the above ground rules and modifications
are:}

\textit{Percentage}\index{Percentage}, which is the \textit{rate}\index{Rate} per hundred or the
\textit{proportion} in one hundred parts. In business
percentage means the duty, interest, or allowance
on a hundred. The word \textit{percentage} is derived
from the Latin \textit{per centum}\index{Per centum}, \textit{per} meaning \textit{by} and
\textit{centum} meaning a \textit{hundred}.

\textit{Interest}\index{Interest} is the per cent of money paid for the
use of money borrowed or otherwise obtained.
The interest to be paid may be either agreed upon
or is determined by the statutes of a state.

\textit{Simple interest}\index{Interest, simple}\index{Simple interest}
is the per cent to be paid a creditor
for the time that the principal remains unpaid
and is usually calculated on a yearly basis, a year
being taken to have $12$ months, of $30$ days each,
or $360$ days.

\textit{Compound interest}\index{Interest, compound}\index{Compound interest}
is the interest on the principal
and the interest that remains unpaid; the
new interest is reckoned on the combined amounts,
when they are considered as a new principal.

\textit{Profit and Loss}\index{Profit and loss}\index{Loss}
are the amounts gained and
the amounts lost when taken together in a business
transaction.
%-----File: 022.png-----

\textit{Gross Profit}\index{Gross profit}\index{Profit, gross}
is the total amount received from
the sale of goods without deductions of any kind
over and above the cost of purchase or production.

\textit{Net Profit}\index{Net profit}\index{Profit, net}
is the amount remaining after all
expenses such as interest, insurance, transportation,
etc., are deducted from the gross cost.

\textit{Loss}\index{Loss} is the difference between the gross profit
and the net profit.

\textit{Reduction of Weights and Measures}%
\index{Reduction of weights and measures}\index{Weights and measures, reduction of}
means to
change one weight or measure into another weight
or measure. This may be done by increasing or
diminishing varying scales.

\textit{Increased} varying scales\index{Increased varying scales}\index{Varying scales, increased}
of weight run thus:
$1$ ounce, $1$ pound, and $1$ ton; for lineal measurement,
$1$ inch, $1$ foot, $1$ yard, etc.; for liquid measurement,
$1$ pint, $1$ quart, $1$ gallon, etc. \textit{Decreased}
varying scales\index{Decreased varying scales}\index{Varying scales, decreased}
run $\frac{1}{2}$ ounce, $\frac{1}{3}$ pound, and $\frac{3}{4}$ ton;
$\frac{1}{4}$ inch; $\frac{1}{2}$ foot, $\frac{3}{4}$ yard, etc.; and $\frac{1}{3}$ pint, $\frac{3}{4}$ quart, and
$\frac{1}{2}$ gallon.
%-----File: 023.png-----

\nbchapter{II}{Rapid Addition}

\begin{nbchaptercontentsone}
Learning to Add Rapidly.\\
The Addition Table.\\
Three Line Exercise.\\
Quick Single Column Addition.\\
Simultaneous Double Column Addition.\\
Left-Handed Two-Number Addition.\\
Simultaneous Three Column Addition.\\
Bookkeepers Check Addition.\\
Adding Backwards (Check Addition).\\
Addition with Periods.\\
To Check Added Work.\\
Lightning Addition (so-called).
\end{nbchaptercontentsone}


\textsc{While} there are no direct short cuts to addition%
\index{Addition, rapid}\index{Rapid addition}
unless one uses an \textit{arithmometer}\index{Arithmometer} or other adding
machine there is an easy way to \textit{learn} to add, and
once learned it will not only make you quick at
figuring but it will aid you wonderfully in other
calculations.

The method by which this can be done is very
simple and if you will spend a quarter or half an
hour a day on it for a month you will be amazed
to find with what speed and ease and accuracy you
will be able to add up any ordinary column of
numbers.

This method is to learn the \textit{addition table} just
as you learned the multiplication table when you
%-----File: 024.png-----
were at school; that is, so that you instantly know
the sum of any two figures below ten just as you
instantly know what the product is of the same
figures.

In ordinary school work children are not taught
%-----File: 025.png-----
the addition table thoroughly and for this reason
very few of them ever become expert at figures.
On the other hand when one takes a course in
some business college the first thing he is given
to do is to learn the addition table.

\nbparagraph{Learning to Add Rapidly.}---There are only 45
combinations that can be formed with the nine
figures and cipher, as the following table shows, %[**TN: original wording "preceding table"]
and these must be learned by \textit{heart}.

\begin{center}
\textit{Addition Table}\index{Addition table}
\small
\[
\begin{array}{*{9}{@{}c@{}c@{}}}
1 &\qquad&1  &\qquad&1  &\qquad&1  &\qquad&1  &\qquad&1  &\qquad&1  &\qquad&1  &\qquad&1\\[\nbskipadjA]
1 &&2  &&3  &&4  &&5  &&6  &&7  &&8  &&9\\[\nbskipadjB]
\cline{1-1}\cline{3-3}\cline{5-5}\cline{7-7}\cline{9-9}\cline{11-11}\cline{13-13}\cline{15-15}\cline{17-17}
2 &&3  &&4  &&5  &&6  &&7  &&8  &&9 &&10\\[\nbskipadjC]
%
  &&2  &&2  &&2  &&2  &&2  &&2  &&2  &&2\\[\nbskipadjA]
  &&2  &&3  &&4  &&5  &&6  &&7  &&8  &&9\\[\nbskipadjB]
\cline{3-3}\cline{5-5}\cline{7-7}\cline{9-9}\cline{11-11}\cline{13-13}\cline{15-15}\cline{17-17}
  &&4  &&5  &&6  &&7  &&8  &&9 &&10 &&11\\[\nbskipadjC]
%
  &&   &&3  &&3  &&3  &&3  &&3  &&3  &&3\\[\nbskipadjA]
  &&   &&3  &&4  &&5  &&6  &&7  &&8  &&9\\[\nbskipadjB]
\cline{5-5}\cline{7-7}\cline{9-9}\cline{11-11}\cline{13-13}\cline{15-15}\cline{17-17}
  &&   &&6  &&7  &&8  &&9 &&10 &&11 &&12\\[\nbskipadjC]
%
  &&   &&   &&4  &&4  &&4  &&4  &&4  &&4\\[\nbskipadjA]
  &&   &&   &&4  &&5  &&6  &&7  &&8  &&9\\[\nbskipadjB]
\cline{7-7}\cline{9-9}\cline{11-11}\cline{13-13}\cline{15-15}\cline{17-17}
  &&   &&   &&8  &&9 &&10 &&11 &&12 &&13\\[\nbskipadjC]
%
  &&   &&   &&   &&5  &&5  &&5  &&5  &&5\\[\nbskipadjA]
  &&   &&   &&   &&5  &&6  &&7  &&8  &&9\\[\nbskipadjB]
\cline{9-9}\cline{11-11}\cline{13-13}\cline{15-15}\cline{17-17}
  &&   &&   &&  &&10 &&11 &&12 &&13 &&14\\[\nbskipadjC]
%
  &&   &&   &&   &&   &&6  &&6  &&6  &&6\\[\nbskipadjA]
  &&   &&   &&   &&   &&6  &&7  &&8  &&9\\[\nbskipadjB]
\cline{11-11}\cline{13-13}\cline{15-15}\cline{17-17}
  &&   &&   &&   &&  &&12 &&13 &&14 &&15\\[\nbskipadjC]
%
  &&   &&   &&   &&   &&   &&7  &&7  &&7\\[\nbskipadjA]
  &&   &&   &&   &&   &&   &&7  &&8  &&9\\[\nbskipadjB]
\cline{13-13}\cline{15-15}\cline{17-17}
  &&   &&   &&   &&   &&  &&14 &&15 &&16\\[\nbskipadjC]
%
  &&   &&   &&   &&   &&   &&   &&8  &&8\\[\nbskipadjA]
  &&   &&   &&   &&   &&   &&   &&8  &&9\\[\nbskipadjB]
\cline{15-15}\cline{17-17}
  &&   &&   &&   &&   &&   &&  &&16 &&17\\[\nbskipadjC]
%
  &&   &&   &&   &&   &&   &&   &&   &&9\\[\nbskipadjA]
  &&   &&   &&   &&   &&   &&   &&   &&9\\[\nbskipadjB]
\cline{17-17}
\ZZ&&\ZZ&&\ZZ&&\ZZ&&  &&   &&   &&  &&18
\end{array}
\]
\end{center}

After learning the addition table so thoroughly
you can skip around and know the sum of any of
the two-figure combinations in the preceding table,
then make up exercises in which three figures
form a column and practice on these until you can
read the sums offhand, thus:

\begin{center}
\textit{Three Line Exercise}\index{Addition, three line exercise}
\small
\[
\begin{array}{*{9}{@{}c@{}c@{}}}
  1  &\qquad&2&\qquad&3&\qquad&4&\qquad&5&\qquad&6&\qquad&7&\qquad&8&\qquad&9\\[\nbskipadjA]
  2  &&      3  &&    4  &&    5  &&    6  &&    7  &&    8  &&    9  &&    1\\[\nbskipadjA]
  3  &&      4  &&    5  &&    6  &&    7  &&    8  &&    9  &&    1  &&    2\\[\nbskipadjB]
\cline{1-1}\cline{3-3}\cline{5-5}\cline{7-7}\cline{9-9}\cline{11-11}\cline{13-13}\cline{15-15}\cline{17-17}
\,6\,  &&  \,9\,&&   12 &&    15 &&    18 &&    21 &&    24 &&    18 &&    12
\end{array}
\]
\end{center}

When you have mastered the three line combinations
as given in the above exercise so that the
instant you glance at any column you can give the
sum of it without having to add the figures repeatedly,
you should extend the exercise to include
all possible combinations of three figures.

Follow this with exercises of four line figures
and when you can do these without mental effort
you are in a fair way to become a rapid calculator
and an accurate accountant.
%-----File: 026.png-----

\nbparagraph{Quick Single Column Addition.}%
\index{Quick single column addition}\index{Addition, quick single column}%
---One of the
quickest and most accurate ways of adding long
columns of figures is to add each column by itself
and write down the unit of the sum under the
units column, the tens of the sum under the hundreds %[**TN: wording as in original text]
column of the example, etc. The two sums
are then added together, which gives the total
sum, as shown in the example on the right.
{\small
\[
\setlength\arraycolsep{0.2em}
\begin{array}{@{}lrrrrl} \\
                       &    (d)& (b)& (c)& (a)&\\
                       &      9&   8&   7&   4&\\
                       &      8&   7&   6&   5&\\
                       &      3&   4&   2&   5&\\
                       &      8&   2&   6&   7&\\
                       &      5&   2&   7&   9&\\
                       &      8&   7&   4&   2&\\\cline{2-5}
\text{Third col.}      &      3&   0&   3&   2&\quad\text{first column}\\
\text{Fourth col.\quad}&4\quad1&   3&   2&    &\quad\text{second column}\\\cline{2-5}
\text{Total}           &4\quad4&   3&   5&   2&\\ \\ \\
\end{array}
\qquad
\begin{array}{l@{}r@{}}
                             & 38\\
                             & 41\\
                             & 23\\
                             & 18\\
                             & 34\\
                             & 16\\
                             & 22\\
                             & 55\\
                             & 66\\
                             & 91\\\cline{2-2}
\text{First column sum}      & 44\\
\text{Second column sum\quad}&36\Z\\\cline{2-2}
\text{Total sum}             &404
\end{array}
\]
}% end \small

Where a column of three, four, or more figures,
as shown in the example on the left of this page, is %[**TN: original wording "on the right"]
to be added, the separate columns are added up in
the order shown by the letters in parenthesis
above each column; that is, the units (a)~column
is added first and the sum set down as usual;
the hundreds (b)~column is added next and the
unit figure of this sum is set under the hundreds
column, which makes 3032; add the tens column
next and set down the units figure of this sum
%-----File: 027.png-----
under the tens column and set the tens figure of
this sum under the hundreds column and finally
add up the thousands column, which makes 4132.
Finally add the two sums together and the total
will be the sum wanted.

There are several reasons why this method of
adding each column separately is better than the
usual method of adding and carrying to the next
column, and among these are (a)~it does away with
the mentally carried number; (b)~a mistake is
much more readily seen; and (c)~the correction is
confined to the sum of the column where the mistake
occurred, and this greatly simplifies the
operation.

\nbparagraph{Simultaneous Double Column Addition.}%
\index{Addition, simultaneous double column}\index{Simultaneous double column addition}%
---While
accountants as a rule add one column of figures at
a time as just described, many become so expert it
is as easy for them to add two columns at the same
time as it is one and besides it is considerably
quicker.

A good way to become an adept at adding two
columns at once is to begin by adding a unit
number to a double column number, thus:
\[
\begin{array}{@{}c@{\quad}r@{}c@{\hspace{6em}}c@{\quad}r@{}}
  (a) &    16 && (b) &      44\\
      &     9 &&     &       8\\\cline{2-2}\cline{5-5}
&          25 &&&           52
\end{array}
\]
The sums of any such combinations should be
known the instant you see them and without the
slightest hesitancy or thought.
%-----File: 028.png-----

Since the sum of the two unit figures have already
been learned and as the tens figure of the units
sum is never more than $1$ it is easy to think, say,
or write $1$ more to the sum of the tens column than
the figure in the tens column calls for.

For instance, $1$ in the tens column of the larger
number ($16$ in the ($9$) case) calls for $2$ in the tens
column of the sum; $3$ in the tens column of the
larger number calls for $4$ in the tens column of
the sum, etc.

Where double columns are added, as
\[
\begin{array}{@{}r@{}}
68\\
87\\\hline
155
\end{array}
\]
the mind's eye sees the sum of both the units
column and the tens column at practically the
same instant, thus:
\[
\begin{array}{@{}r@{}}
68\\
87\\\hline
141\llap{\raisebox{-6pt}{\rotatebox[origin=c]{90}{\{}}}5\\[-0.3ex] %[**TN: \underbrace is too wide]
5\:\Z
\end{array}
\]
carries the tens figure of the sum of the units
column, which in this case is $1$, and adds it to the
units figure of the second sum, which in this case
is $4$, the total sum, which is $155$, is had.

After you are sufficiently expert to rapidly add
up single columns it is only a step in advance and
one which is easily acquired, especially where the
sum is not more than $100$, to add up two columns
%-----File: 029.png-----
at the same time, and following this achievement
summing up three or more numbers is as easily
learned.

{\nbstretchyspace
\nbparagraph{Left-Handed Two-Number Addition.}%
\index{Addition, left-handed, two number}\index{Left-handed two number addition}%
---This is
a somewhat harder method of addition than the
usual one but it often affords relief to an overworked
mind to change methods and what is
more to the point it affords an excellent practice
drill.}

Take as an example
\[
\begin{array}{@{}r@{}}
 492\\
 548\\\hline
1040\\
\end{array}
\]

Begin by adding the left-hand or hundreds
column first, then add the tens column, and finally
the units column; in the above example $5 + 4 = 9$ %[**TN: original wording "5 + 4 = 10"]
and a glance will show that $1$ is to be carried, hence
put down $10$ in your grey matter for the sum on
the left-hand side; $4 + 9 = 13$ and another glance
at the units column shows that $1$ is also to be
carried and $1$ added to $3 = 4$, so put down $4$ for the
tens column of the sum, and as $8 + 2 = 10$ and the
$1$ having already been carried put down $0$ in the
units column of the sum.

Left-hand addition is just like saying the alphabet
backward---it is as much of a novelty and
far more useful---in that it is just as easy to do
it as the right-hand frontward way---after it is
once learned.
%-----File: 030.png-----

\nbparagraph{Simultaneous Three Column Addition.}%
\index{Addition, simultaneous three column}\index{Simultaneous three column addition}---

% The Simultaneous Three Column Addition example contains lines indicating
% which digits are placed in which locations. Since pdfTeX with the eepic
% package generates low-quality lines, an image has been used instead.
% The image p23.pdf was created using p23.tex, which in turn was generated
% using ePiX from p23.xp. These source files may be found in a subfolder
% of the images folder.
\nbsubparagraph{Example.}---
\begin{center}\includegraphics[width=4in]{images/p23.pdf}\end{center}

\nbsubparagraph{Rule.}---An easy and rapid way to add three
columns of figures at the same time is to take the
upper number ($541$) and add the units figure of
the next lower number ($7$) to it ($548$); then make
the tens figure ($3$) a multiple of $10$ (in this case it
is $30$) and add it to the first sum ($578$); %[**TN: original wording "548"]
 then make
the hundreds figure a multiple of $100$ (in this it is
$200$) and add it to the last sum ($778$) and so on
with each figure of each number to the last
column, when the total will be the sum of the
column.

When adding three columns of numbers by this
method you can start with the lower number and
add upward just as well as starting with the upper
number and adding down. By adding mentally,
many of the operations which are shown in the
right-hand column do not appear, only the succeeding
sums being noted.
%-----File: 031.png-----

This operation is especially useful in figuring
up cost sums, thus:
\[
\begin{tabular}{@{}r@{}}
\$\Z.75\\
   4.20\\
   1.95\\
   2.37\\
    .25\\\hline
 \$9.52
\end{tabular}
\]

Beginning with the lower amount we have
25---32---62---262---267---357---457---477---877---882---and 9.52.

\nbparagraph{Bookkeepers Check Addition.}%
\index{Addition, bookkeeper's check}\index{Bookkeeper's check addition}\index{Check addition, bookkeeper's}%
---This method
is largely used by bookkeepers and others where
there are apt to be interruptions since it is simple,
single column addition, and as the sum of each
column is set down by itself there is no carrying,
hence there is small chance for errors and it can
be easily checked up.

The rule is to put down separately the sum of
the units column on the top right-hand side, the
sum of the tens column with the unit figure under
the tens figure of the first added, and so on until
all of the columns have been added and their
individual sums arranged in the order given,
thus:
%-----File: 032.png-----
\[
\begin{tabular}[b]{@{}rrrrr@{}}
 5&  7&  6&  4&  2\\
 7&  3&  2&  5&  4\\
 5&  3&  6&  6&  4\\
 3&  9&  4&  3&  5\\
 5&  2&  7&  1&  3\\
 1&  6&  8&  4&  6\\
 4&  3&  6&  9&  2\\
\hline
\textbf{30}& \textbf{33}& \textbf{39}& \textbf{32}& \textbf{26}\\
33&  7\rlap{,}& 2&  4&  6\\
\end{tabular}
\qquad\qquad\qquad
\begin{tabular}[b]{@{}l@{}}
\,\textbf{\ZZZZ26}\\
\,\textbf{\ZZZ32}\\
\,\textbf{\ZZ39}\\
\,\textbf{\Z33}\\
\,\textbf{30}\\
\hline
\textbf{337,246}
\end{tabular}
\]

\nbparagraph{Adding Backward (Check Addition).}%
\index{Adding Backward}---The sums
of each of the above columns may be set down in
the reverse order to that shown above; that is,
with the sum of the unit column at the lower right-hand
side, the sum of the tens column with the
unit figure over the tens figure of the last sum,
and so on until all of the sums of the separate
columns have been written down in this order:
\[
\begin{tabular}{l}
 \textbf{30}\\
\textbf{\Z33}\\
\textbf{\ZZ39}\\
\textbf{\ZZZ32}\\
\textbf{\ZZZZ26}\\
\hline
 \textbf{337246}
\end{tabular}
\]

\nbparagraph{Period Addition.}%
\index{Addition, period}\index{Period addition}%
---A great help in adding up
long single columns is to use periods to mark off
10's; that is, the units are added up until the one
is reached where the sum is just less than 20; this
one is checked off with a period, or other mark
which stands for 10; the amount of the sum over
%-----File: 033.png-----
10 is carried and added to the next unit and the
adding goes on until the sum is again just less
than 20, when this figure is checked off, and so on
to the top of the column; the last sum is either
mentally noted, or it can be written down and
then added to the tens as indicated by the periods,
when the total will be the sum of the single
column.
\[
{\setlength\arraycolsep{2pt}
\begin{array}{@{}rclccl@{}}
7 & &(14)                 & & & 14\\
4 &.&(17 \text{ carry }7) &.&=& 10\\ %[**TN: original wording ". = 1.0"]
6 & &                     & & &\\
3 &.&(17 \text{ carry }7) &.&=& 10\\
5 & &                     & & &\\
8 &.&(19 \text{ carry }9) &.&=& 10\\
7 & &                     & & &\\
1 &.&(14 \text{ carry }4) &.&=& 10\\
6 & &                     & & &\\
2 &.&(17 \text{ carry }7) &.&=& 10\\
4 & &                     & & &\\
3 & &                     & & &\\
8 & &                     & & &\\
\hline
64& &                     & & & 64
\end{array}}
\]

\nbparagraph{To Check Addition.}%
\index{Addition, to check}\index{Check addition, to}%
---To ascertain whether or
not the work that has been added is correct it
should be checked up by some one of the various
methods given in connection with the above examples.
As good a method as any is to add each
column from the bottom up and then from the
top down.
%-----File: 034.png-----

\nbparagraph{Lightning Addition.}%
\index{Addition, lightning}\index{Lightning addition}%
---The rules given in the
preceding pages cover practically all of the real
helps in making addition a quick, easy, and accurate
operation. To add any number of figures on
sight is a delusion and a snare, a trick pure and
simple, which you or any one can do when you
know the secret, and as such it will be explained
under the caption of \textit{The Magic of Figures} in
\hyperref[chapter:IX]{Chapter~IX} of this book. %[**TN: original wording "Chapter VIII"]
%-----File: 035.png-----

\nbchapter{III}{Rapid Subtraction}
\index{Rapid subtraction}\index{Subtraction, rapid}

\begin{nbchaptercontentsone}
The Taking-Away Method.\\
The Subtraction Table.\\
Subtraction by Addition.\\
Combined Addition and Subtraction.\\
Subtracting Two or More Numbers from Two or More Other Numbers.\\
To Check the Work.
\end{nbchaptercontentsone}

\textsc{It} has been previously pointed out that addition
and subtraction are \textit{universal} operations and hence
they are closely related.

There are two methods in use by which the
difference or \textit{remainder} between two numbers can
be found and these are (1)~the \textit{taking-away}, or
\textit{complement} method, and (2)~the \textit{making-up}, or
\textit{making-change} method, as these methods are variously
called.

\nbparagraph{The Taking-Away Method.}%
\index{Complement in subtraction}\index{Taking-away method of subtraction}\index{Subtraction, taking-away method}%
---In the taking-away
or complement the difference or remainder
between two numbers is found by \textit{thinking down}\index{Thinking down in subtraction}
from the whole number, or \textit{minuend}\index{Minuend}, to the smaller
number, or \textit{subtrahend}\index{Subtrahend}.

The difference between two numbers, or remainder,
is called the \textit{complement} for the simple reason
that it completes what the subtrahend lacks to
make up the minuend; thus in subtracting $4$ from
$9$ the remainder is $5$ and hence $5$ is the complement
of $4$.
%-----File: 036.png-----

When subtraction is performed by the taking
away, or complement method, the subtraction
table should be thoroughly learned, and as subtraction
is the inverse operation of addition, of course
but $45$ combinations can be made with the nine
figures and the cipher and these are given in the
following table.

This table should be learned so that the remainder
of any of the two-figure combinations
given in the above table can be instantly named,
and this is a very much easier thing to do than to
learn the addition, since the largest remainder is $9$; and %[**TN: 'largest' added]
when the table is learned letter perfect, rapid
subtraction becomes a simple matter.

\begin{center}
\textit{The Subtraction Table}\index{Subtraction table}
\small
\[
\begin{array}{*{9}{@{}c@{}c@{}}}
2	&\qquad&	&\qquad&	&\qquad&	&\qquad&	&\qquad&	&\qquad&	&\qquad&	&\qquad&\\[\nbskipadjA]
1	&&	&&	&&	&&	&&	&&	&&	&&\\[\nbskipadjB]
\cline{1-1}
1	&&	&&	&&	&&	&&	&&	&&	&&\\[\nbskipadjC]
%
3	&&3	&&	&&	&&	&&	&&	&&	&&\\[\nbskipadjA]
1	&&2	&&	&&	&&	&&	&&	&&	&&\\[\nbskipadjB]
\cline{1-1}\cline{3-3}
2	&&1	&&	&&	&&	&&	&&	&&	&&\\[\nbskipadjC]
%
4	&&4	&&4	&&	&&	&&	&&	&&	&&\\[\nbskipadjA]
1	&&2	&&3	&&	&&	&&	&&	&&	&&\\[\nbskipadjB]
\cline{1-1}\cline{3-3}\cline{5-5}
3	&&2	&&1	&&	&&	&&	&&	&&	&&\\[\nbskipadjC]
%
5	&&5	&&5	&&5	&&	&&	&&	&&	&&\\[\nbskipadjA]
1	&&2	&&3	&&4	&&	&&	&&	&&	&&\\[\nbskipadjB]
\cline{1-1}\cline{3-3}\cline{5-5}\cline{7-7}
4	&&3	&&2	&&1	&&	&&	&&	&&	&&\\[\nbskipadjC]
%
6	&&6	&&6	&&6	&&6	&&	&&	&&	&&\\[\nbskipadjA]
1	&&2	&&3	&&4	&&5	&&	&&	&&	&&\\[\nbskipadjB]
\cline{1-1}\cline{3-3}\cline{5-5}\cline{7-7}\cline{9-9}
5	&&4	&&3	&&2	&&1	&&	&&	&&	&&\\[\nbskipadjC]
%
7	&&7	&&7	&&7	&&7	&&7	&&	&&	&&\\[\nbskipadjA]
1	&&2	&&3	&&4	&&5	&&6	&&	&&	&&\\[\nbskipadjB]
\cline{1-1}\cline{3-3}\cline{5-5}\cline{7-7}\cline{9-9}\cline{11-11}
6	&&5	&&4	&&3 &&2	&&1	&&	&&	&&\\[\nbskipadjC]
%
8	&&8	&&8	&&8	&&8	&&8	&&8	&&	&&\\[\nbskipadjA]
1	&&2	&&3	&&4	&&5	&&6	&&7	&&	&&\\[\nbskipadjB]
\cline{1-1}\cline{3-3}\cline{5-5}\cline{7-7}\cline{9-9}\cline{11-11}\cline{13-13}
7	&&6	&&5	&&4	&&3	&&2	&&1	&&	&&\\[\nbskipadjC]
%
9	&&9	&&9	&&9	&&9	&&9	&&9	&&9	&&\\[\nbskipadjA]
1	&&2	&&3	&&4	&&5	&&6	&&7	&&8	&&\\[\nbskipadjB]
\cline{1-1}\cline{3-3}\cline{5-5}\cline{7-7}\cline{9-9}\cline{11-11}\cline{13-13}\cline{15-15}
8	&&7	&&6	&&5	&&4	&&3	&&2	&&1	&&\\[\nbskipadjC]
%
% \llap used to avoid making column two digits wide, hence \cline can be used for horizontal lines
\llap{1}0&&\llap{1}0	&&\llap{1}0	&&\llap{1}0	&&\llap{1}0	&&\llap{1}0	&&\llap{1}0	&&\llap{1}0	&&\llap{1}0\\[\nbskipadjA]
1	&&2	&&3	&&4	&&5	&&6	&&7	&&8	&&9\\[\nbskipadjB]
\cline{1-1}\cline{3-3}\cline{5-5}\cline{7-7}\cline{9-9}\cline{11-11}\cline{13-13}\cline{15-15}\cline{17-17}
9	&&8	&&7	&&6	&&5	&&4	&&3	&&2	&&1\\
\end{array}
\]
\end{center}
%-----File: 037.png-----

\nbparagraph{Subtraction by Addition.}%
\index{Austrian method of subtraction}\index{Making change, method of subtraction}%
\index{Making-up method of subtraction}\index{Subtraction by addition}%
\index{Subtraction, Austrian method}\index{Subtraction, making change method}%
\index{Subtraction, making-up method}
---If the subtraction
table has never been thoroughly learned, then the
easiest way to do subtraction is by the \textit{making-up},
or \textit{making-change} method, and which is also sometimes
called the \textit{Austrian} method. This method
is very simple in that it consists of \textit{adding} to the
subtrahend a number large enough to equal the
minuend.

For example, suppose a customer has bought an
article for $22$ cents and he hands the clerk a $\$1.00$
bill. In giving him his change the clerk hands
him $3$ pennies and says ``$25$''; then he hands him
a quarter and says ``$50$,'' and finally he hands
him a half-dollar and says ``and $50$ makes $\$1.00$.''

He has performed the operation of subtracting
$22$ cents from $\$1.00$ by simple addition, since $22$
cents added to $3$ pennies make $25$ cents, and the
quarter added to the $25$ cents makes $50$ cents and
the half-dollar added to the $50$ cents makes up the
dollar, or worked out in simple figures it becomes
\begin{align*}
\phantom{\text{subtrahend\;\;}\nearrow\;} 22 + \underbrace{3 + 25 + 50}
                  &= \$1.00\\[-1.5ex]
         \text{subtrahend\;\;}\raisebox{0.5ex}{$\nearrow$\;}\phantom{ 22 + 3 + 25 + 50}
                  &\phantom{{}= \$1.00}\raisebox{0.5ex}{$\nwarrow$\;\;} \text{minuend}\\[-1ex]
\phantom{\text{subtrahend\;\;}\nearrow\; 22 + 3 +}.78\rlap{\:$\leftarrow$ \text{remainder}} \phantom{{}+ 50} &
\end{align*}
%-----File: 038.png-----
or to do it by the complement method, then
\[
\begin{array}{@{}r@{}l}
\$1.00 & \text{ amount tendered}\\
   .22 & \text{ amount purchased}\\\cline{1-1}
   .78 & \text{ difference returned in change.}
\end{array}
\]

\nbparagraph{Combined Addition and Subtraction.}%
\index{Addition and subtraction combined}\index{Combined addition and subtraction}%
---In business
calculations it is often necessary to add two
or more numbers and then subtract the sum of
them from another number. When this is the case
the operations of addition and subtraction need
not be performed separately but they may be
done in one operation.

Suppose, by way of illustration, that you have a
certain balance in the bank to your credit and you
have made, let us say, four checks against it, and
that you want to know what the balance is after the
amounts of the checks have been deducted, thus:
\[
\begin{array}{@{}l@{}c@{}l}
% \left. \right. to balance the \left. \right\} later
\left.\$462.76\right.&& \text{original balance}\\\cline{1-1}
\multicolumn{2}{@{}l@{\;}}{
\left.\begin{array}{@{}l@{}}
\phantom{\$12}3.80\\
\phantom{\$1}16.50\\
\phantom{\$1}12.16\\
\phantom{\$12}7.69
\end{array}\;\right\}}&\text{checks drawn}\\\cline{1-1}
\left.\phantom{\$}422.61\right.&&\text{new balance} %[**TN: text alignment adjusted]
\end{array}
\]

In this operation each column of the checks drawn
is added and the unit of the sum is subtracted from
the corresponding column of the amount of the
original balance and the remainder is put down
under this column for the new balance.

In the above example, for instance, the $6$ and
%-----File: 039.png-----
$9$ of the units column of the checks drawn are
added, making $15$, and the $5$ of the latter number
is subtracted from the $6$ of the units column of
the original balance, and $1$ is put down in the
units column of the new balance.

The $1$ from the $15$ is carried and added to the
tens column of the checks drawn, which makes $21$,
and the $1$ of the latter is subtracted from the $7$
in the tens column of the original balance and the
remainder, $6$, is put down in the tens column of
the new balance.

Next the $2$ of the added tens column is carried
and added to the hundreds column of the checks
drawn, which makes $20$, and the cipher subtracted
from the $2$ of the original balance leaves $2$, which
is put down in the hundreds column of the new
balance.

Finally the $2$ of the added hundreds column is
carried and added to the thousands column, making
$4$, and this subtracted from $6$ in the thousands
column of the original balance leaves $2$, which is
set down in the thousands column of the new
balance, and putting down the remaining $4$ of the
original balance a new balance is had of $\$422.61$.

\nbparagraph{Subtracting Two or More Numbers from Two
or More Other Numbers.}%
\index{Subtracting two or more numbers from two or more other numbers}%
---There are often cases
where two or more numbers have to be subtracted
from two or more other numbers, and this can be
easily and quickly done by an extension of the
method just given.
%-----File: 040.png-----

Suppose you have made deposits for each day
in the week excluding Sunday, and on each day
of the week you have made certain checks; as
an example, say that your deposits and checks
have run for a week like this:
\[
\begin{array}{l@{}r@{}c@{}l}
\begin{array}{l}
\text{Mon.}\\
\text{Tues.}\\
\text{Wed.}\\
\text{Thurs.}\\
\text{Fri.}\\
\text{Sat.}
\end{array}&
\multicolumn{2}{l}{
\left.
\begin{array}{@{}r@{}}
\$27.82 \\
33.46\\
17.25\\
48.21\\
19.25\\
35.45\\\cline{1-1}
\end{array}
\;\right\}}&
\text{Deposits}\\
\begin{array}{l}
\text{Mon.}\\ \\ \\ \\ \\ \\
\end{array}&
\multicolumn{2}{l}{
\left.
\begin{array}{@{}r@{}}
\$10.20 \\
  15.35\\
  32.01\\
   7.16\\
  11.25\\
  35.45\\\cline{1-1}
\end{array}
\;\right\}}&
\text{Checks Made}\\
&
\multicolumn{2}{l}{
\left.
\begin{array}{@{}r@{}}
70.02\\
181.44\\
111.42\\\cline{1-1}
70.02
\end{array}
\right.}
&
\begin{array}{@{}l@{}}
\text{Balance}\\
\text{total deposits}\\
\text{checks}\\
\text{balance}
\end{array}
\end{array}
\]

In this case add up the units column of the
\textit{deposits first} and then add up the units column of
the checks, subtract the latter from the former
number and put down the units figure, which is
$2$, in the units column; carry the tens figure, which
is $2$, of the units column of the deposits and add
it to the tens column of the deposits. Then carry
the tens figure of the unit column of the checks
%-----File: 041.png-----
drawn, which is $1$, and add it to the tens column
of the deposits and subtract as before, and so on
until all of the columns of both deposits and
checks drawn have been added separately and
subtracted and the remainders put down.

\nbparagraph{To Check the Work.}%
\index{Checking subtraction}\index{Subtraction, checking}%
---To check the results of
subtraction all that is needed is to add the remainder
and the subtrahend together and the sum
of these two numbers will be the same as the
minuend, that is, if the answer is right.
%-----File: 042.png-----

\nbchapter{IV}{Short Cuts in Multiplication}
\index{Multiplication, short cuts in}\index{Short cuts in multiplication}

\begin{center}
\textbf{The Multiplication Table}
\end{center}

\begin{nbchaptercontentsnum}{2em}
\item[(1)] A Further Extension of the Multiplication Table.
\item[(2)] To Find the Square of Two Numbers when Both End in $5$.
\item[(3)] To Find the Product of Two Numbers when Both End in $5$ and the Tens Figures are even.
\item[(4)] To Find the Product of Two Numbers when Both End in $5$ and the Tens Figures are Uneven.
\item[(5)] To Multiply any Number by $10$, $100$, $1000$, etc.
\item[(6)] To Multiply any Number by a Multiple of $10$, as $20$, $300$, $4000$, etc. %[**TN: original wording "Any"]
\item[(7)] To Multiply any Number Ending in Ciphers by a Multiple of $10$.
\item[(8)] To Multiply any Number by $25$ or $75$.
\item[(9)] To Multiply any Number by Higher Multiples of $25$, as $125$, $250$, etc.
\item[(10)] To Square any Number Formed of Nines. %[**TN: original wording "Any"]
\item[(11)] To Multiply a Number by $11$ or any Multiple of $11$, as $22$, $33$, $44$, etc. %[**TN: original wording "Any"]
\item[(12)] To Multiply a Number by $21$, $31$, $41$, etc.
\item[(13)] To Multiply a Number by $101$, $201$, $301$, etc.
\item[(14)] To Square a Number Having Two Figures.
\item[(15)] To Multiply any Two Numbers of Two Figures Each where the Units are Alike.
\item[(16)] To Multiply any Two Numbers of Two Figures Each where the Tens are Alike.
\item[(17)] To Multiply any Number by a Number Formed of Factors.
\item[(18)] The Complement Method of Multiplying.
\item[(19)] The Supplement Method of Multiplying.
\item[(20)] The Sliding Method of Multiplying.
\end{nbchaptercontentsnum}

{\footnotesize
To Check the Work:
\begin{nbchaptercontentscheck}
\item[Check\quad I.]---By Division
          \item[II.]---By Division
         \item[III.]---By Casting Out Nines
          \item[IV.]---Lightning Method.
\end{nbchaptercontentscheck}
}%end footnotesize
%-----File: 043.png-----

To become an adept at short cuts in arithmetic
the multiplication table must be so well learned
that the product of any two numbers up to and
including $12$ may be instantly expressed without
thinking.

After the table is learned---indeed there are very
few who are not already proficient---the next step
is to know the short-cut rules and how to form
certain number combinations, and then---\emph{practice}
until you have mastered them.

When these principles are followed you will be
able to solve a large number of arithmetical problems
with lightninglike rapidity. %[**TN: original wording "lightning-like" - changed for consistency]

\begin{center}
\textit{The Multiplication Table}\index{Multiplication table}
\small
\[
\setlength\arraycolsep{2pt}
\renewcommand\arraystretch{1.5}
\begin{array}{|*{16}{c|}}
\hline
\textbf{\Z}\textbf{1}\textbf{\Z} & \textbf{\Z}2\textbf{\Z} &
\textbf{\Z}3\textbf{\Z}          & \textbf{\Z}4\textbf{\Z} &
\textbf{\Z}5\textbf{\Z}          & \textbf{\Z}6\textbf{\Z} &
\textbf{\Z}7\textbf{\Z}          & \textbf{\Z}8\textbf{\Z} &
\textbf{\Z}9\textbf{\Z} & 10 &11 &12 &13 &14 &15 &\textbf{\Z}\textbf{1}\textbf{\Z}\\
\hline
\textbf{2} & \textbf{4} & 6 & 8 & 10 &12 &14 &16 &18 &20 &22 &24 &26 &28 &30 &\textbf{2}\\
\hline
\textbf{3} & 6 & \textbf{9} & 12 &15 &18 &21 &24 &27 &30 &33 &36 &39 &42 &45 &\textbf{3}\\
\hline
\textbf{4} & 8 & 12 &\textbf{16} &20 &24 &28 &32 &36 &40 &44 &48 &52 &56 &60 &\textbf{4}\\
\hline
\textbf{5} & 10 &15 &20 &\textbf{25}
\smash{\rlap{\hspace{5pt}\llap{\rule[-7pt]{111pt}{2pt}}\llap{\rule[-7pt]{2pt}{104pt}}}} &
30 &35 &40 &45 &50 &55 &60 &65 &70 &75 &\textbf{5}\\
\hline
\textbf{6} & 12 &18 &24 &30 &\textbf{36} &42 &48 &54 &60 &66 &72 &78 &84 &90 &\textbf{6}\\
\hline
\textbf{7} & 14 &21 &28 &35 &42 &\textbf{49} &56 &63 &70 &77 &84 &91 &98 &105 & \textbf{7}\\
\hline
\textbf{8} & 16 &24 &32 &40 &48 &56 &\textbf{64} &72 &80 &88 &96 &104& 112& 120 &\textbf{8}\\
\hline
\textbf{9} & 18 &27 &36 &45 &54 &63 &72 &\textbf{81} &90 &99 &108& 117& 126& 135 &\textbf{9}\\
\hline
\textbf{10} &20 &30 &40 &50 &60 &70 &80 &90 &\textbf{100}
\smash{\rlap{\hspace{3pt}\llap{\rule[-7pt]{222pt}{2pt}}\llap{\rule[-7pt]{2pt}{208pt}}}} &
110& 120& 130& 140& 150 &\textbf{10}\\
\hline
\textbf{11} &22 &33 &44 &55 &66 &77 &88 &99 &110& \textbf{121}& 132& 143& 154& 165 &\textbf{11}\\
\hline
\textbf{12} &24 &36 &48 &60 &72 &84 &96 &108& 120& 132& \textbf{144}& 156& 168& 180 &\textbf{12}\\
\hline
\textbf{13} &26 &39 &52 &65 &78 &91 &104& 117& 130& 143& 156& \textbf{169}& 182& 195 &\textbf{13}\\
\hline
\textbf{14} &28 &42 &56 &70 &84 &98 &112& 126& 140& 154& 168& 182& \textbf{196}& 210 &\textbf{14}\\
\hline
\textbf{15} &30 &45 &60 &75 &90 &105& 120& 135& 150& 165& 180& 195& 210& \textbf{225} &\textbf{15}\\
\hline
\end{array}
\]
\end{center}
%-----File: 044.png-----

\bigskip
Having mastered the multiplication table you
are provided with a mental master-key which will
quickly unlock any of the following short-cut rules.

\nbparagraph{(1) A Further Extension of the Multiplication
Table.}---\textit{Multiplying by the Teens}.%
\index{Multiplication table, a further extension of the}%
\index{Multiplying by the teens}%[**TN: original wording "tens"]
\begin{nbalignexample}
\nbexample{18 × 15 &= 270 \nbanswer}
\end{nbalignexample}

\nbsubparagraph{Rule.}---To the multiplicand add the unit figure
of the multiplier and annex a cipher; to this add
the product obtained by multiplying the unit
figures of both the multiplier and the multiplicand,
thus:
\begin{align*}
 18 + 5  &= \Z23\text{ or annexing a cipher}\\
         &= 230\\
  5 × 8  &= \Z40\text{ or}\\
230 + 40 &= 270 \nbanswer
\end{align*}

\nbparagraph{(2) To Find the Square of Two Numbers when
Both End in 5.}%
\index{Finding the square of two numbers when both end in 5}%
\index{Multiplication, to find the square of two numbers when both end in 5}
\begin{nbalignexample}
\nbexample{25 × 25 &= 625}
\end{nbalignexample}

\nbsubparagraph{Rule.}---\nblabel{a}~Multiply the 5's in the units column
and write down 25 for the ending figures of the
product; \nblabel{b}~add 1 to the first 2 (or other figure)
in the tens column which makes 3 and then \nblabel{c}~multiply
the 3 by the 2 in the other tens column
thus:
\begin{align*}
25 ×  25 &= \\
 5 × \Z5 &= \Z25\\
 2 × \Z3 &= 6\\[\nboverlineskip]
         &\E\nboverline{625} \nbanswer
\end{align*}
%-----File: 045.png-----
\nbparagraph{(3) To Find the Product of Two Numbers when
Both End in 5 and the Tens Figures are Even.}%
\index{Finding the product of two numbers when both end in 5 and the tens figures are even}%[**TN: original wording "the tens are even"]
\index{Multiplication, to find the product of numbers when both end in 5 and the tens figures are even}
\begin{nbalignexample}
\nbexample{25 × 45 &= 1125}
\end{nbalignexample}

\nbsubparagraph{Rule.}---\nblabel{a}~Multiply the 5's in the units column
as before and write down 25 for the ending
figures of the product; \nblabel{b}~add the figures of the
tens column (in the above example $2 + 4 = 6$)
and \nblabel{c}~divide the sum (6) by 2; \nblabel{d}~multiply the
two figures in the tens column ($2 × 4 = 8$) and
\nblabel{e}~add the remainder ($6 ÷ 2 = 3$) which will
give the last figure of the product (11) thus:
\begin{align*}
                  25 ×  45 & =\\
\nblabel{a}\quad \Z5 × \Z5 & = \ZZZ25\\
\nblabel{b}\quad \Z2 + \Z4 & =  6\\
\nblabel{c}\quad \Z6 ÷ \Z2 & =  3\\
\nblabel{d}\quad \Z2 × \Z4 & =  8\\
\nblabel{e}\quad \Z3 + \Z8 & = \Z11\\[\nboverlineskip]
                           &\E \Z\nboverline{1125} \nbanswer
\end{align*}

\nbparagraph{(4) To Find the Product of Two Numbers when
Both End in 5 and the Tens Figures are Uneven.}%
\index{Finding the product of two numbers when both end in 5 and the tens figures are uneven}%
\index{Multiplication, to find the product of two numbers when both end in 5 and the tens figures are uneven}
\begin{nbalignexample}
\nbexample{35 × 65 &= 2275}
\end{nbalignexample}

\nbsubparagraph{Rule.}---\nblabel{a}~Where the tens figures are odd (as
3 and 6) write down 75 for the ending figures of
the product; \nblabel{b}~add the tens figures ($3 + 6 = 9$),
\nblabel{c}~divide the sum (9) by 2 (which gives $4 \frac{1}{2}$), \nblabel{d}~leave
off the fraction and add the quotient (4)
to \nblabel{e}~the product found by multiplying the figures
of the tens columns ($3 × 6 = 18$) and write the
result (22) before the 75, thus:
%-----File: 046.png-----
\begin{align*}
                  35 ×  65 & =  2275\\
\nblabel{a}\quad \Z5 × \Z5 & \text{ write down }75\\
\nblabel{b}\quad \Z3 + \Z6 & = \Z9\\
\nblabel{c}\quad \Z9 ÷ \Z2 & = \Z4 \tfrac{1}{2}\\
\nblabel{d}\quad \Z3 × \Z6 & =  18\\
\nblabel{e}\quad \Z4 +  18 & = \qquad 22\\[\nboverlineskip]
                           &\E \qquad\nboverline{2275}
\end{align*}

\nbparagraph{(5) To Multiply any Number by 10, 100, 1000, etc.}%
\index{Multiplication, to multiply any number by 10, 100, 1000, etc.}%
\index{Multiplying any number by 10, 100, 1000, etc.}%
---This is the simplest of all short methods
of multiplication.
\begin{nbalignexample}
\nbexamples{824 &× \ZZ10}\\
         && 824 &× \Z100 \\
         && 824 &×  1000
\end{nbalignexample}

\nbsubparagraph{Rule.}---Annex as many ciphers as there are
ciphers in the multiplier, thus:
\begin{align*}
&824 × 1\mathit{0}    \qquad=\qquad   824\mathit{0}\\
&\phantom{824 × 10}   \raisebox{0.5ex}{$\nwarrow$}\; \text{one cipher}    \;\raisebox{0.5ex}{$\nearrow$}\\
&824 × 1\mathit{00}   \qquad=\qquad   824\mathit{00}\\
&\phantom{824 × 10\;}  \raisebox{0.5ex}{$\nwarrow$}\; \text{two ciphers}   \;\raisebox{0.5ex}{$\nearrow$}\\
&824 × 1\mathit{000}  \qquad=\qquad   824\mathit{000}\\
&\phantom{824 × 10\;\;} \raisebox{0.5ex}{$\nwarrow$}\; \text{three ciphers} \;\raisebox{0.5ex}{$\nearrow$}
\end{align*}

\nbparagraph{(6) To Multiply any Number by a Multiple of 10, as 20, 300, 4000, etc.}%[**TN: original wording "Any"]
\index{Multiplication, to multiply any number by a multiple of 10}%
\index{Multiplying any number by a multiple of 10}%
\begin{nbalignexample}
\nbexamples{49 &× \ZZ20}\\
        && 732 &× \Z300\\
        && 841 &×  2000
\end{nbalignexample}

\nbsubparagraph{Rule.}---Multiply the multiplicand by the first
figure of the multiplier and annex as many ciphers
%-----File: 047.png-----
to the product as there are ciphers in the multipliers,
thus:
\[
\setlength\arraycolsep{3em}
\begin{array}{lll}
            49            &    \phantom{219,6}\llap{732}  &    \phantom{1,682}\llap{841}\\
           \Z2            &    \phantom{219,6}\llap{3}    &    \phantom{1,682}\llap{2}\\
\nboverline{98}\mathbf{0} & \nboverline{219,6}\mathbf{00} & \nboverline{1,682,}\mathbf{\,000}
\end{array}
\]

\nbparagraph{(7) To Multiply any Number Ending in Ciphers
by a Multiple of 10.}%
\index{Multiplication, to multiply any number ending in ciphers by a multiple of 10}%
\index{Multiplying any number ending in ciphers by a multiple of 10}
\begin{nbalignexample}
\nbexamples{240 × \phantom{,12}20 &= \phantom{,1234}4,800}\\
       && 3,200 × \phantom{,1}600 &= \phantom{1}1,920,000&&\\
       && 5,000 ×           3,000 &=           15,000,000&&
\end{nbalignexample}

\nbsubparagraph{Rule.}---Disregard the ciphers in the multiplicand
and multiply the figures of it by the figures
of the multiplier and annex as many ciphers to the
product as there are ciphers in the multiplicand
and the multiplier together, thus:
\[
\setlength\arraycolsep{3em}
\begin{array}{lll}
          24               &   \phantom{,1}32                &            \Z5\\
         \Z2               &   \phantom{,12}6                &            \Z3\\
\nboverline{48}\mathbf{00} & \nboverline{1,92}\mathbf{0,000} & \nboverline{15}\mathbf{,000,000}
\end{array}
\]

\nbparagraph{(8) To Multiply any Number by 25 or 75.}%
\index{Multiplication, to multiply any number by 25 or 75}%
\index{Multiplying any number by 25 or 75}%
---In
this short cut the desired product is obtained by
\textit{division}.
\begin{nbalignexample}
\nbexamples{\nblabel{a}\quad 219 × 25 &= 5,475}\\
         && \nblabel{b}\quad 523 × 75 &= 4,025
\end{nbalignexample}

\nbsubparagraph{Rule A.}---First be it known that
\begin{flalign*}
          && 100 ÷ 25 &= 4\\
\text{and}&& 100 ÷ 75 &= \tfrac{4}{3} && \phantom{\text{and}} %[**TN: original wording "\tfrac{3}{4}"]
\end{flalign*}

In the above examples annex two ciphers to the
multiplicand, thus:
%-----File: 048.png-----
\begin{align*}
\nblabel{a}\quad& 21900\\
\nblabel{b}\quad& 52300
\end{align*}

Now in example \nblabel{a} since $100 ÷ 25 = 4$, by
dividing $21900$ by $4$ gives the same result as multiplying
$219$ by $25$, thus:
\[
\begin{array}{lr@{\;}r@{}l}
\nblabel{a}& 4 &\begin{array}{|@{\;}r@{}} 21900 \\\hline\end{array}&\\
           &   &                           5475 &\nbanswer
\end{array}
\]

\nbsubparagraph{Rule B.}---Add two ciphers to the multiplicand
as explained in Rule~\textit{A}, and since $75$ is $\frac{3}{4}$ of $100$
divide the multiplicand by $4$ and then multiply
the quotient by $3$, thus:
\[
\begin{array}{r@{\;}r@{}l}
4 &\begin{array}{|@{\;}r@{}} 52300\\ \hline \end{array}&\\
  &                          13075                     &\\
  &                              3                     &\\
  &\begin{array}{@{\;}r@{}} \hline 39225 \end{array}   &\nbanswer
\end{array}
\]

\nbparagraph{(9) To Multiply any Number by Higher Multiples of 25.}%[**TN: original wording "Any"]
\index{Multiplication, to multiply any number by higher multiples of 25}%
\index{Multiplying any number by higher multiples of 25}
\begin{nbalignexample}
\nbexamples{\nblabel{a}\quad 334 &× 125}\\
         && \nblabel{b}\quad 642 &× 250
\end{nbalignexample}

\nbsubparagraph{Rule.}---Where multiples of $25$ larger than $75$
are used as the multipliers three ciphers are annexed
to the multiplicand, and the multiplier is
divided by $1000$, thus in example \nblabel{a}
\[
1000 ÷ 125 = 8
\]
and by dividing the multiplicand by 8 we get
\[
\begin{array}{lr@{\;}r@{}l}
\nblabel{a}& 8 &\begin{array}{|@{\;}r@{}} 334,000 \\\hline\end{array}&\\
   &   &                           41,750                    &\nbanswer
\end{array}
\]
%-----File: 049.png-----

In example \nblabel{b}
\[
1000 ÷ 250 = 4
\]
and by the same operation we have
\[
\begin{array}{lr@{\;}r@{}l}
\nblabel{b} %[**TN: original label "(a)"]
& 4 &\begin{array}{|@{\;}r@{}} 642,000 \\\hline\end{array}&\\
   &   &                          160,500                  &\nbanswer
\end{array}
\]

\nbparagraphdisplay{(10) To Square any Number Formed of Nines.}
\begin{nbalignexample}
\nbexample{999&=998001}\\
      && 9,999&=99980001 \text{\quad(Answers)}&\\
      &&99,999&=9999800001                    &
\end{nbalignexample}

\nbsubparagraph{Rule.}---Put down from left to right as many
nines less one as the number to be squared contains,
one $8$, as many ciphers as there are nines in
the answer and annex a $1$.

\nbparagraph{(11) To Multiply a Number by 11 or any Multiple of 11.}%[**TN: original wording "Any"]
\index{Multiplication, to multiply a number by 11, or any multiple of 11}%
\index{Multiplying a number by 11 or any multiple of 11}
\begin{nbalignexample}
\nbexamples{\nblabel{a}\quad 592×11 &=  6,512}\\
          &&\nblabel{b}\quad 643×33 &= 21,219
\end{nbalignexample}

\nbsubparagraph{Rule A.}---Where $11$ is used as the multiplier,
write down the multiplicand and under it write
down the multiplicand over again but with the
units column under the tens column of the first
number, and then add the two numbers, thus:
\[
\begin{array}{ll@{}l}
\nblabel{a} &           \Z592\\
            &            592 \\
            &\nboverline{6512} &\nbanswer
\end{array}
\]

\nbsubparagraph{Rule B.}---Where a higher multiple of $11$ is
used, as the multiplier of $33$, $44$, $66$, write the
number of the multiplicand down twice as explained
%-----File: 050.png-----
in Rule~\textit{A}, add them together and then
multiply the sum by $3$, $4$, $6$, or other unit of the
multiplier as the case may be, thus:
\[
\begin{array}{ll@{}l}
\nblabel{b} &            \ZZ643  &\\
            &            \Z643   &\\
            &\Z\nboverline{7073} &\\
            &            \ZZZZ3  &\\
            & \nboverline{21219} &\nbanswer
\end{array}
\]

\nbparagraph{(12) To Multiply a Number by 21, 31, 41, etc.}%
\index{Multiplication, to multiply a number by 21, 31, 41, etc.}%
\index{Multiplying a number by 21, 31, 41, etc.}
\begin{nbalignexample}
\nbexample{342 × 81 &= 27,702}
\end{nbalignexample}

\nbsubparagraph{Rule.}---\nblabel{a}~First write down the unit figure of
the multiplicand (which in this case is $2$) for the
unit figure of the answer; \nblabel{b}~multiply the unit
figure of the multiplicand by the tens figure of
the multiplier ($8×2=16$); \nblabel{c}~add the product
($16$) to the tens figure of the multiplicand ($4+16=20$);
\nblabel{d}~set down the unit figure of the sum ($0$)
in the answer and carry the tens figure ($2$); \nblabel{e}~then
multiply the tens figures of both multiplicand
and multiplier ($8×4=32$); \nblabel{f}~add now the tens
figure of the sum carried ($2$) to the product above
($32+2=34$).

\nblabel{g}~Add the hundreds in the multiplicand ($3$)
to the above sum ($34+3=37$); \nblabel{h}~and set down
the unit figure of this sum ($7$) in the answer and
carry the tens figure ($3$); \nblabel{i}~multiply the hundreds
figure of the multiplicand ($3$) by the tens figure
of the multiplier ($8×3=24$); \nblabel{j}~add the figure
carried ($3+24=27$), and as this is the last operation
%-----File: 051.png-----
\nblabel{k}~set down the sum ($27$) before the first
figures of the product and this will be the answers.

While this may seem a long rule it is quite simple
and once put into actual practice any number can
be multiplied by any number less than $100$, if it
ends in~$1$, with great rapidity.

Following the rule step by step in figures it works
thus:
\begin{nbalignsteps}
\nblabel{a}\quad & \text{write down } 2 \text{ for the answer}\\*
\nblabel{b}\quad & 8 × 2 = 16\\
\nblabel{c}\quad & 4 + 16 = 20\\
\nblabel{d}\quad & \text{write down } 0 \text{ for the answer}\\
\nblabel{e}\quad & 8 × 4 = 32\\
\nblabel{f}\quad & 32 + 2 = 34\\
\nblabel{g}\quad & 34 + 3 = 37\\
\nblabel{h}\quad & \text{write down } 7 \text{ for the answer}\\
\nblabel{i}\quad & 8 × 3 = 24\\
\nblabel{j}\quad & 3 + 24 = 27\\
\nblabel{k}\quad & \text{write down } 27 \text{ for the answer}\\*
                 & \phantom{\text{write down } 27 \text{ for the }} 27702 \rlap{\nbanswer}
\end{nbalignsteps}

\nbparagraph{(13) To Multiply a Number by 101, 301, 501, etc.}%
\index{Multiplication, to multiply a number by 101, 301, 501, etc.}%
\index{Multiplying a number by 101, 301, 501, etc.}
\begin{nbalignexample}
\nbexample{536 × 401 &= 214936}
\end{nbalignexample}

\nbsubparagraph{Rule.}---\nblabel{a}~Put down the unit and tens figures
of the multiplicand ($36$) for the answer; \nblabel{b}~multiply
the unit figure of the multiplicand and the
hundreds figure of the multiplier ($6 × 4 = 24$);
\nblabel{c}~add the sum ($24$) and the hundreds figure of
the multiplicand ($24 + 5 = 29$); \nblabel{d}~put down the
unit figure of this sum ($9$) and carry the tens
%-----File: 052.png-----
figure (2); \nblabel{e}~multiply the tens and hundreds
figures of the multiplicand (53) by the hundreds
figure of the multiplier ($4 × 53 = 212$); \nblabel{f}~to
this sum (212) add the carried figure ($212 + 2
= 214$); \nblabel{g}~write down the last sum (214) in
the result and this will be the answer (214,936).

The rule works out in figures thus:
\begin{nbalignsteps}
\nblabel{a}\quad & \text{put down } 36 \text{ for the answer}\\*
\nblabel{b}\quad & 6 × 4 = 24\\
\nblabel{c}\quad & 24 + 5 = 29\\ %[**TN: original wording "24 × 5 = 29"]
\nblabel{d}\quad & \text{put down } 9 \text{ and carry } 2\\
\nblabel{e}\quad & 4 × 53 = 212\\
\nblabel{f}\quad & 212 + 2 = 214\\
\nblabel{g}\quad & \text{write down } 214 \text{ for the answer}\\*
                 & \phantom{\text{write down } 214 \text{ for the }} 214,936 \nbanswer
\end{nbalignsteps}

\nbparagraph{(14) To Square a Number Having Two Figures.}%
\index{Multiplication, to square a number having two figures}%
\index{Squaring a number having two figures}
\begin{nbalignexample}
\nbexample{73 × 73 &= 5329}
\end{nbalignexample}

\nbsubparagraph{Rule A}.---\nblabel{a}~Multiply the unit figure of the
multiplicand ($3$) by the unit figure of the multiplier
($3 × 3 = 9$) and write down the product;
\nblabel{b}~add the unit figure of the multiplicand ($3$)
and that of the multiplier ($3 + 3 = 6$); \nblabel{c}~multiply
one of the tens figures ($7$) by the last sum
($7 × 6 = 42$); \nblabel{d}~write down the unit of the
product ($2$) and carry the tens ($4$); \nblabel{e}~multiply
the tens figures together ($7 × 7 = 49$); \nblabel{f}~and add
the figure carried ($49 + 4 = 53$) and write down
this last sum ($53$) for the answer, thus:\\
\hspace*{\fill}$\mathbf{5329}$\hspace*{\fill}\\
or
%-----File: 053.png-----
\begin{nbalignsteps}
\nblabel{a}\quad & 3 × 3 = \mathbf{9}\\*
                 & \text{write down the } \mathbf{9}\\
\nblabel{b}\quad & 3 + 3 = 6\\
\nblabel{c}\quad & 7 × 6 = 42\\ %[**TN: original wording "7 + 6"]
\nblabel{d}\quad & \text{put down } \mathbf{2} \text{ and carry } 4\\
\nblabel{e}\quad & 7 × 7 = 49\\ %[**TN: original wording "7 + 7"]
\nblabel{f}\quad & 49 + 4 = 53\\*
                 & \text{write down } \mathbf{53}
\end{nbalignsteps}
%
\begin{nbalignexample}
\nbexample{76 × 76 &= 5772}
\end{nbalignexample}

\nbsubparagraph{Rule.}---Where a product larger than $9$ results
when the units of numbers are multiplied, and a
figure remains to be carried this rule applies:

\nblabel{a}~Multiply the unit of the multiplicand (6) by
the unit of the multiplier ($6 × 6 = 36$); \nblabel{b}~write
down the unit of this product ($6$) for the answer,
and carry the tens figure ($3$); \nblabel{c}~add the tens
figures of the multiplicand and multiplier ($7 + 7 =
14$); \nblabel{d}~multiply the sum ($14$) by one of the
units ($14 × 6 = 84$), \nblabel{e}~and to the latter product
($84$) add the carried figure ($84 + 3 = 87$); \nblabel{f}~write
down the unit figure of this sum ($7$) for
the answer, and carry the tens figure ($8$); \nblabel{g}~multiply
the tens figures ($7 × 7 = 49$); \nblabel{h}~add
the carried figure ($49 + 8 = 57$), \nblabel{i}~and write
down this sum ($57$) and the whole number will
be the square of the numbers multiplied ($5776$).

The rule may be simplified thus:
\begin{nbalignsteps}
\nblabel{a}\quad & 6 × 6 = 36\\*
\nblabel{b}\quad & \text{write down } \mathbf{6} \text{ and carry } 3\\
%-----File: 054.png-----
\nblabel{c}\quad & 7 + 7 = 14\\
\nblabel{d}%[**TN: original label "(g)"]
\quad & 14 × 6 = 84\\
\nblabel{e}\quad & 84 + 3 = 87\\
\nblabel{f}\quad & \text{write down } \mathbf{7} \text{ and carry } 8\\
\nblabel{g}\quad & 7 × 7 = 49\\
\nblabel{h}\quad & 49 + 8 = 57\\*
\nblabel{i}\quad & \text{write down } \mathbf{57}
\end{nbalignsteps}

\nbparagraph{(15) To Multiply any Two Numbers of Two
Figures Each Where the Units are Alike.}%
\index{Multiplication, to multiply any two numbers of two figures each where the units are alike}%
\index{Multiplying any two numbers of two figures each where the units are alike}
\begin{nbalignexample}
\nbexample{53 × 63 &= 3339}
\end{nbalignexample}

\nbsubparagraph{Rule.}---\nblabel{a}~Multiply the unit figures ($3 × 3
= 9$); \nblabel{b}~and write down the product ($9$) for
the answer; \nblabel{c}~add the tens figures ($5 + 6 = 11$);
\nblabel{d}~multiply the sum ($11$) by one of the units
($11 × 3 = 33$); \nblabel{e}~write down the unit of this
sum ($3$) for the answer and carry the tens figure (3);
\nblabel{f}~multiply the tens figures ($5 × 6 = 30$); \nblabel{g}~add
this product ($30$) to the carried figure ($30 + 3
= 33$), \nblabel{h}~and write down this sum ($33$), when the
whole number will be the answer ($3339$); or simplified
it becomes:
\begin{nbalignsteps}
\nblabel{a}\quad & 3 × 3=9\\*
\nblabel{b}\quad & \text{write down the } \mathbf{9}\\
\nblabel{c}\quad & 5 + 6 = 11\\
\nblabel{d}\quad & 11 × 3 = 33\\
\nblabel{e}\quad & \text{write down } \mathbf{3} \text{ and carry } 3\\
\nblabel{f}\quad & 5 × 6 = 30\\
\nblabel{g}\quad & 30 + 3 = 33\\*
\nblabel{h}\quad & \text{write down } \mathbf{33}
\end{nbalignsteps}
%-----File: 055.png-----

\nbparagraph{(16) To Multiply any Two Numbers of Two
Figures Each Where the Tens are Alike.}%
\index{Multiplication, to multiply any two numbers of two figures each where the tens are alike}%
\index{Multiplying any two numbers of two figures each where the tens are alike}
\begin{nbalignexample}
\nbexample{42 × 45 &= 1890}
\end{nbalignexample}

\nbsubparagraph{Rule.}---\nblabel{a}~Multiply the unit figures $(5 × 2
= 10)$; \nblabel{b}~write down the unit of the sum (0)
and carry the tens figure (1); \nblabel{c}~add the unit
figures ($5 + 2 = 7$); \nblabel{d}~multiply this sum (7) by
one of the tens figures ($4 × 7 = 28$); \nblabel{e}~add the
product (28) to the figure carried ($28 + 1 = 29$);
\nblabel{f}~write down the unit figure of the sum (9)
and carry the tens figure (2); \nblabel{g}~multiply the
tens figures ($4 × 4 = 16$); \nblabel{h}~add the product (16)
to the carried figure ($16 + 2 = 18$), \nblabel{i}~and write
the sum (18) for the answer (1890); or it may be
written thus:
\begin{nbalignsteps}
\nblabel{a}\quad & 5 × 2 = 10\\*
\nblabel{b}\quad & \text{write } \mathbf{0} \text{ and carry } 1\\
\nblabel{c}\quad & 5 + 2 = 7\\
\nblabel{d}\quad & 7 × 4 = 28\\
\nblabel{e}\quad & 28 + 1 = 29\\
\nblabel{f}\quad & \text{write } \mathbf{9} \text{ and carry } 2\\
\nblabel{g}\quad & 4 × 4 = 16\\*
\nblabel{h}\quad & 16 + 2 = \mathbf{18}
\end{nbalignsteps}

\nbparagraph{(17) To Multiply any Number by a Number
Formed of Factors.}%
\index{Multiplication, to multiply any number by a number formed of factors}%
\index{Multiplying any number by a number formed of factors}
\begin{nbalignexample}
\nbexample{384 × 648 &= 248832}
\end{nbalignexample}

\nbsubparagraph{Rule A}.---In this case $64$ can be factored by $8$,
that is $8 × 8 = 64$. \nblabel{a}~Multiply the multiplicand
($384$) by the unit figure of the multiplier ($8 × 384 = 3072$); %[**TN: missing bracket added after 3072]
%-----File: 056.png-----
\nblabel{b}~now since $8$ is a factor of $64$ multiply
the last product ($3072$) by one of the factors
$(3072 × 8 = 24576)$; %[**TN: original wording "3072 × 24576"]
\nblabel{c}~put down this product so
that the unit figure ($6$) will be under the tens
column of the first product ($7$) and add the two
products and the sum will be the product of the
two numbers.

Worked out the operations are:
\begin{align*}
\nblabel{a}
\begin{array}{r}
           384\\
             8\\
\nboverline{3072}
\end{array} &&
\nblabel{b}
\begin{array}{r}
           3072\\
              8\\
\nboverline{24576}
\end{array} &&
\nblabel{c}
\begin{array}{r@{}l}
            3072\\
          24576\Z\\
\nboverline{248832} & \nbanswer
\end{array}
\end{align*}
or
\[
\begin{array}{r@{}l}
               384\\
               648\\
  \nboverline{3072}\\
            24576\Z\\
\nboverline{248832} & \nbanswer
\end{array}
\]
%
\begin{nbalignexample}
\nbexample{485 × 432 &= 209,520}
\end{nbalignexample}

\nbsubparagraph{Rule B.}---In this case $32$ can be factored, since
$4 × 8 = 32$. \nblabel{a}~Multiply the multiplicand ($485$)
by the hundreds figure of the multiplier ($4 × 485 =
1940$); \nblabel{b}~now since $8$ is a factor of $32$ multiply
this product ($1940$) by $8$ which gives $15520$; \nblabel{c}~write
the last product ($15520$) two figures to the
right of the first product ($1940$) and add, when the
sum will be the product wanted, thus:
\begin{align*}
&\nblabel{a}
\begin{array}{r}
             485\\
               4\\ %[**TN: original wording "4 4"]
\nboverline{1940}
\end{array}
&&\nblabel{b}
\begin{array}{r}
             1940\\
                8\\
\nboverline{15520}
\end{array}
&&\nblabel{c}
\begin{array}{r@{}l}
           1940\ZZ\\
            15520\\
\nboverline{209,520} & \nbanswer
\end{array}
\end{align*}
%-----File: 057.png-----
or
\[
\begin{array}[t]{r@{}l}
                485\\
                432\\
 \nboverline{1940\ZZ}\\
              15520\\
\nboverline{209,520}& \nbanswer
\end{array}
\]

\nbparagraph{(18) The Complement Method of Multiplying.}%
\index{Complement method of multiplying}%
\index{Multiplying by the complement method}
\begin{nbalignexample}
\nbexample{93 × 95 &= 8835}
\end{nbalignexample}

\nbsubparagraph{Rule.}---The \textit{complement} of a number is the
figure required to bring it up to $100$ or some multiple
of~$100$; thus $5$ is the complement of $95$ and
$7$ is the complement of~$93$.

\nblabel{a}~Multiply the complements of the two numbers
($7 × 5 = 35$) and \nblabel{b}~write down the product
($35$) for the answer; \nblabel{c}~subtract either complement
from either number %[**TN: should strictly say "from the other number"]
(say $93 - 5 = 88$) and
\nblabel{d}~write down the remainder for the answer ($88$),
thus:
\begin{align*}
&93 × 95 = 8835\\
&\;\phantom{9}7\phantom{3×9}5 \rightarrow\text{complement}\\
&\;\phantom{97}\raisebox{0.5ex}{$\nwarrow$}\text{complement}
\end{align*}
%
\begin{nbalignsteps}
\nblabel{a}\quad & 7 × 5 = \mathbf{35}\\*
\nblabel{b}\quad & \text{write down } 35 \text{ for the answer}\\
\nblabel{c}\quad & 93 - 5 = \mathbf{88}\\
\nblabel{d}\quad & \text{write down } 88 \text{ for the answer}\\*
                 & \rlap{which is}\phantom{\text{write down } 88 \text{ for the }} 8835 \nbanswer
\end{nbalignsteps}

\nbparagraph{(19) The Supplement Method of Multiplying.}%
\index{Multiplying by the supplement method}%
\index{Supplement method of multiplying}
\begin{nbalignexample}
\nbexample{106 × 105 &= 11130}
\end{nbalignexample}

\nbsubparagraph{Rule.}---The supplement of a number is its
unit figure over $100$; thus $6$ is the supplement of
$100$ and $5$ is the supplement of $105$.
%-----File: 058.png-----

\nblabel{a}~Multiply the first two units of the numbers
($6 × 5 = 30$) and \nblabel{b}~write down the product ($30$)
for the last figures of the answer; \nblabel{c}~add the
supplement of either number to the other number
($106 + 5 = 111$), and \nblabel{d}~write down the sum ($111$)
for the first figures of the answer, thus:
\begin{nbalignsteps}
\nblabel{a}\quad & 6 × 5 = 30\\*
\nblabel{b}\quad & \text{write down } 30 \text{ for the answer}\\
\nblabel{c}\quad & 106 + 5 = 111\\
\nblabel{d}\quad & \text{write down } 111 \text{ for the answer}\\*
                 & \rlap{which is}\phantom{\text{write down } 111 \text{ for the }} 11130 \nbanswer
\end{nbalignsteps}

\nbparagraph{(20) The Sliding Method of Multiplying.}%
\index{Multiplying by the sliding method}%
\index{Sliding method of multiplying}
\begin{nbalignexample}
\nbexample{324 × 123 &= 39,852}
\end{nbalignexample}

\nbsubparagraph{Rule.}---To multiply by the sliding method it is
necessary to write the two numbers on separate
slips of paper. As an illustration in the above
example the multiplicand, or larger number, is
written in the reverse order, as $423$ instead of $324$.

Now place the slips of paper one under the
other so that the problem looks like this:
\[
\begin{array}{@{}l@{}}
\ZZ423\\
 123\\
\hline
\end{array}
\]

Next multiply the $4$ by the $3$ thus: $4 × 3 = 12$,
set down the $2$ down in the answer, and carry the
$1$; slide the upper number along until the figures
are in this order:
\[
\begin{array}{l}
\Z423\\
 123\\
\hline
\end{array}
\]
%-----File: 059.png-----

Multiply the 4 by the 2 and the 2 by the 3 thus:
$4 × 2 = 8$ and $2 × 3 = 6$. Add these products and
the 1 which was carried from above thus: $6 + 8 +
1 = 15$. Put down the 5 in the answer and carry
the 1, and slide the upper numbers along until
they bear this relation:
\[
\begin{array}{l}
423\\
123\\
\hline
\end{array}
\]

This time multiply all three sets of numbers,
thus $4 × 1 = 4, 2 × 2 = 4$, and $3 × 3 = 9$. Add
these products and the 1 which was carried from
the above thus: $4 + 4 + 9 + 1 = 18$; write the
8 down in the answer and carry the 1, and slide
the upper slip along until the numbers are in this
position:
\[
\begin{array}{l}
 423\\
\Z123\\
\hline
\end{array}
\]

Multiply the 2 by 1 and the 3 by 2 thus: $2 × 1 = 2$
and $3 × 2 = 6$. To the sum of these products
add the 1 carried from above thus: $2 + 6 + 1 = 9$;
set the 9 down in the answer and again slide the
upper slip until the numbers are thus arranged:
\[
\begin{array}{l}
 423\\
\ZZ123\\
\hline
\end{array}
\]

Finally multiply the 3 by the 1 and write the
product down in the answer and when this is
done we have as the product:
\[
324 × 123 = 39,852 \nbanswer
\]
%-----File: 060.png-----

\nbparagraph{To Check the Work.}%
\index{Checking multiplication work}%
\index{Multiplication, to check the work}%
---There are several ways
to check up products in multiplication but the
following are the shortest and best.

\smallskip
\nbsubparagraph{Check I.---By Division.}%
\index{Checking multiplication by division}%
\index{Multiplication, to check by division}

\nbsubparagraph{Example:} \quad $32 × 58 = 1856$

\nbsubparagraph{Check:} \quad $1856 ÷ 58 = 32$

\nbsubparagraph{Rule.}---Divide the product by the multiplier and
the quotient will be the multiplicand.

\smallskip
\nbsubparagraph{Check II.---By Division.}%
\index{Checking multiplication by division}%
\index{Multiplication, to check by division}

\nbsubparagraph{Example:} \quad $9 × 8 × 2 × 7 × 5 = 5040$

\nbsubparagraph{Check:} \quad $5040 ÷ 9 = 560 = 8 × 2 × 7 × 5$

\nbsubparagraph{Rule.}---Divide the product by one of the factors,
when the quotient should be equal to the product
of the remaining factors.

\smallskip
\nbsubparagraph{Check III.---By Casting out the Nines.}%
\index{Checking multiplication by casting out nines}%
\index{Multiplication, to check by casting out nines}
\nbreflabel{section:nines}
\begin{nbalignexample}
    \nbexample{34  &=            \Z7}\\
           &&  26  &=            \Z8\\
&&\nboverline{204} &\E\nboverline{56} = 2\\
           && 68\Z &\\
&&\nboverline{884} &= 2
\end{nbalignexample}

The example is proved as follows by \textit{casting out}
the nines, that is to take out all the nines contained
in the multiplicand and using the remainders.

\nbsubparagraph{First Step.}---Cast out the nines in the multiplicand
and set down the remainder, or excess of
nines. Since in this case there are three nines in
the multiplicand and $3 × 9 = 27$ then the remainder
will be equal to $34 - 27$ or 7.
%-----File: 061.png-----

\nbsubparagraph{Second Step.}---Cast out the nines in the multiplier
and set down the remainder. There are
two nines in 26 and since $2 × 9 = 18$ the remainder
is equal to $26 - 18$ or 8.

\nbsubparagraph{Third Step.}---Multiply the remainders by each
other: In the above example $7 × 8 = 56$, and cast
the nines out of the product thus obtained. In
this case there are 6 nines in 56 and since $9 × 6
= 54$ the remainder $= 56 - 54$ or 2.

\nbsubparagraph{Fourth Step.}---Now cast the nines out of the
answer. In this case there are 98 nines in the
answer and since $9 × 98 = 882$ then the remainder
is $884 - 882$ or 2. And since the product of the
excesses of nine is 56 and the excess of nines in
56 is 2 and the excess of nines in the answer is 2
the work is proved correct.

\nbsubparagraph{Check IV.\quad The Lightning Method.}%
\index{Checking multiplication by the lightning method}%
\index{Lightning method of checking multiplication}%
\index{Multiplication, lightning method of checking}%
---A simple
way of proving multiplication with lightninglike
rapidity is based on casting out the nines. It is also
called the method of \textit{unitates}. Now the unitate
of a number is the sum of the figures which
make up the number reduced to one figure, thus:
\begin{center}
\begin{tabular}{lllllll}
\multicolumn{2}{l}{number} &
\multicolumn{2}{l}{sum}    &
\multicolumn{2}{l}{sum}    &
unitate\\
       &&      && \multicolumn{2}{l}{reduced}\\
347653 &=& 28 &=& 10 &    =& 1\\
503792 &=& 26 &=&    &     & 8\\
437965 &=& 34 &= &   &     & 7
\end{tabular}
\end{center}
The sum of the figures in 347653 is 28; by adding
the figures in 28 we have 10 and adding the figures
in 10 we get 1.
%-----File: 062.png-----

The way in which the method of unitates is
applied to checking multiplication is as follows:

\nbsubparagraph{Example 1:}
\[
\begin{array}{l@{~}r}
  \left.
  \begin{array}{r@{~}c@{~}r}
    \ZZZ432 & = & \Z9\\
    \ZZZ331 & = & \Z7
  \end{array}
  \right\}
  &
  63 = 9
  \\
  \left.
  \begin{array}{r@{~}c@{~}r}
    \nboverline{432}\\
              1296\Z\\
             1296\ZZ\\
    \phantom{142992}\llap{$\nboverline{142,992}$} &=& 27
  \end{array}\right.
  &
  \begin{array}{l@{}}
    \phantom{\nboverline{1}}\\ \\ \\ \phantom{\nboverline{1}} = 9
  \end{array}
\end{array}
\]

In this example the unitate of the multiplicand
(432) is 9 and the unitate of the multiplier (331)
is 7. Now multiplying these unitates we have
$9 × 7 = 63$ and the unitate of 63 is 9.

An examination of the product shows that the
sum of the figures comprising it is 27 and the unitate
of 27 is 9. Therefore since the unitate of the
product, multiplier, and the multiplicand is 9 also,
the work must be correct.

\nbsubparagraph{Example 2:}
\[
\begin{array}{l@{~}r}
  \left.
  \begin{array}{r@{~}c@{~}c@{~}c@{~}r}
    \ZZ459\Z & = & 18 & = & \Z9\\
    \ZZZ230  & = &    &   & \Z5
  \end{array}
  \right\}
  &
  = 45 = 9
  \\
  \left.
  \begin{array}{r@{~}c@{~}c@{~}c@{~}r}
    \nboverline{13770}\\
                918\ZZ\\
    \phantom{105570}\llap{$\nboverline{105,570}$} & = & \ZZ & \E & 18
  \end{array}\right.
  &
  \begin{array}{l@{}}
    \phantom{\nboverline{1}}\\ \\ \phantom{\nboverline{1}} = 9
  \end{array}
\end{array}
\]

The simplicity of this check makes it easily used
and hence it should be done mentally.
%-----File: 063.png-----

\nbchapter{V}{Short Cuts in Division}
\index{Division, short cuts in}
\index{Short cuts in division}

\begin{nbchaptercontentsone}
To Instantly Find the Divisor of a Number.\\
To Divide any Number by 10, 100, 1000, etc.\\ %[**TN: original wording "Any"]
To Divide a Number by 25 or a Multiple of 25, as 50, 75, 125, etc.\\ %[**TN: original wording "as 5, 75, 125"]
\noindent \hangindent=4em To Divide a Number by a Number Formed of Factors (where there are no Remainders).\\
\noindent \hangindent=4em To Divide a Number by a Number Formed of Factors (where there is a Remainder).\\
To Divide a Number by a Number Formed of Three Factors.\\
To Check the Work:\\
\qquad \phantom{I}I.\quad By Casting out the Nines.\\
\qquad           II.\quad The Lightning Method.
\end{nbchaptercontentsone}

\textsc{While} there are no tables to learn as an aid to
short-cut division there are however a few rules
which, if you know them, will instantly show what
figure can be used as a divisor and hence much
time and effort is saved in trying out different
figures to find a divisor.

\nbparagraphdisplay{To Instantly Find the Divisor of a Number.}%
\index{Division, to instantly find the division of a number}%
\index{Instantly finding the divisor of a number}

(\textit{A}) An even number can, of course, always be
divided by \textbf{2}, thus:
\[
32574
\]

(\textit{B}) When the sum of the figures forming a
number can be divided by \textbf{3} the number itself
can be divided by \textbf{3}, thus:
\[
651\quad (12÷ 3=4)
\]
%-----File: 064.png-----

(\textit{C}) When a number ends in two ciphers, or the
last two numbers can be divided by \textbf{4} then the
whole number can be divided by \textbf{4}, thus:

$3200$ or $868$, because in the latter case $68$ can be
divided by \textbf{4}.

(\textit{D}) When a number ends with either a cipher
or \textbf{5} then the number can be divided by \textbf{5}, thus:
\[
3200\text{ or }775
\]

(\textit{E}) When the number is even and the sum of
the figures forming it can be divided by \textbf{3} then the
number can be divided by \textbf{6}, thus:
\[
1284 \qquad (1 + 2 + 8 + 4 = 15\text{ and }15 ÷ 3 = 5)
\]

%[**TN: Rule F is not generally true! The note referred to (after the index)
%       attempts to explain it, but divisibility by 7 cannot be determined
%       solely from the sums of odd and even place figures]
(\textit{F})\nbreflabel{section:rulef} When the sum of the odd place figures of a
number minus the sum of the even place figures
can be divided by \textbf{7} then the whole number can
be divided by \textbf{7}, thus:
\[
\begin{array}{cl}
\phantom{\text{even place}}\quad\raisebox{-0.5ex}{$\swarrow$}\text{even place}\\
\overbrace{\hspace{2.2em}}\\[-0.5em]
\parbox{5em}{$2 \hfill 2 \hfill 7 \hfill 2 \hfill 9$}& (18 - 4 = 14)\\[-1em]
\underbrace{\hspace{2.2em}}\underbrace{\hspace{2.2em}}\\
\parbox{5em}{$\phantom{2} \hfill \uparrow \hfill \phantom{7} \hfill \uparrow \hfill \phantom{9}$}\\
\text{odd place}&\text{\footnotesize* See note on page~\pageref{section:notef}.}
\end{array}
\]

(\textit{G}) When the last three place figures of a number
can be divided by \textbf{8} then the whole number can
be divided by \textbf{8}, thus:
\[
54184 \qquad\qquad (184 ÷ 8 = 23)
\]

(\textit{H}) When the sum of all the figures can be
divided by \textbf{9} then the whole number can be divided
by \textbf{9}, thus:
\[
33498 \qquad (3 + 3 + 4 + 9 + 8 = 27\text{ and }27 ÷ 9 = 3)
\]
%-----File: 065.png-----

%[**TN: factors formatted bold in paragraphs I to K for consistency with A to H]
(\textit{I}) When the last figure of a number is a cipher
it can, of course, always be divided by \textbf{10}, thus:
\[
39870
\]

(\textit{J}) When the sum of the odd place figures minus
the sum of the even place figures of a number can
be divided by \textbf{11}, then the whole number can be
divided by \textbf{11}, thus:
\[
\begin{array}{c}
\phantom{\text{even place}}\quad\raisebox{-0.5ex}{$\swarrow$}\text{even place}\\
\overbrace{\hspace{2.2em}}\\[-0.5em]
\parbox{5em}{$8 \hfill 6 \hfill 4 \hfill 1 \hfill 6$}\\[-1em]
\underbrace{\hspace{2.2em}}\underbrace{\hspace{2.2em}}\\
\parbox{5em}{$\phantom{8} \hfill \uparrow \hfill \phantom{4} \hfill \uparrow \hfill \phantom{6}$}\\
\text{odd place}\\
\end{array}
\begin{array}{l}
(8 + 4 + 6 = 18;\; 6 + 1 = 7;\\
18 - 7 = 11 \text{ and } 11 ÷ 11 = 0) %[**TN: Since 11 ÷ 11 = 1, assume this refers to zero remainder]
\end{array}
\]

(\textit{K}) When the \textit{sum} of the figures of a number
can be divided by three \textit{and} each of the last two
figures of the number can be divided by four,
then the number can be divided by \textbf{12}.
\[
4104 \qquad (4 + 1 + 0 + 4 = 9;\; 9 ÷ 3 = 3 \text{ and } 4 ÷ 4 = 0) %[**TN: Since 4 ÷ 4 = 1, assume this refers to zero remainder]
\]

\nbparagraph{(1) To Divide any Number by 10, 100, 1000, etc.}%[**TN: original wording "Any"]
\index{Dividing any number by 10, 100, 1000, etc.}%
\index{Division, to divide any number by 10, 100, 1000, etc.}%
\begin{nbalignexample}
\nbexample{3722 ÷ 10 &= 372\tfrac{2}{10}}
\end{nbalignexample}

\nbsubparagraph{Rule.}---To divide any number by 10 or by a
multiple of 10 it is only necessary to put a decimal
point before the last figure, thus:
\begin{flalign*}
         &&&372.2\\
\text{or}&&&372\tfrac{2}{10}&&\phantom{\text{or}}
\end{flalign*}

When a number is divided by $100$ point off two
figures with the decimal point; when dividing by
$1000$ point off three figures with the decimal point,
etc.
%-----File: 066.png-----

\nbparagraph{(2) To Divide a Number by 25 or a Multiple of
25, as 50, 75, 125, etc.}%[**TN: original wording "as 5, 75, 125"]
\index{Division, to divide a number by 25 or a Multiple of 25}%
\index{Dividing a number by 25, or a multiple of 25}%
\begin{nbalignexample}
\nbexample{4600 ÷ 25 &= 184}
\end{nbalignexample}

\nbsubparagraph{Rule A.}---As $25$ is $\frac14$ of $100$ drop the ciphers of the
dividend ($4600$) and multiply the remaining figures
($46$) by $4$, thus:
\[
46 × 4 = 184 \nbanswer
\]

\nbsubparagraph{Rule B.}---As $50$ is $\frac12$ of $100$ drop the ciphers of the %[**TN: added em-dash for consistency with Rule A]
dividend and multiply by $2$; $75$ is $\frac34$ of $100$---drop
the ciphers of the dividend, multiply the remaining
figures by $4$ and divide by $3$; $125$ is $\frac54$ of $100$---drop
the ciphers of the dividend, multiply the
remaining figures by $4$ and divide the product
by $5$.

\nbparagraph{(3) To Divide any Number by a Number Formed of Factors.}%[**TN: original wording "Any"]
\index{Dividing any number by a number joined by factors}%
\index{Division, to divide any number by a number formed by factors; where there are no remainders}%
\index{Dividing any number by a number formed of factors; where there are no remainders}%
---\nbsubparagraph{Where there are no remainders.}
\begin{nbalignexample}
\nbexample{12775 ÷ 35 &= 365}
\end{nbalignexample}

\nbsubparagraph{Rule.}---$5 × 7 = 35$, hence $5$ and $7$ are factors
of $35$. Divide the dividend ($12775$) by one of the
factors ($5$) and divide the quotient ($2555$) by the
other factor ($7$) and this quotient ($365$) will be
the answer. In other words the example is done
by two operations of short division instead of one
operation of long division, thus:
\[
\begin{array}{r@{}r@{}r@{}l}
5\;\vline &\multicolumn{2}{@{}r@{}}{\,12775}\\\cline{2-3}
          &\llap{7}\;\vline &2555\\\cline{3-3}
&&365&\nbanswer
\end{array}
\]
%-----File: 067.png-----

\nbparagraph{(4) To Divide any Number by a Number Formed of Factors.}%[**TN: original wording "Any"]
\index{Dividing any number by a number formed of factors; where there is a remainder}%
\index{Division, to divide any number by a number formed by factors; where there is a remainder}%
---\nbsubparagraph{Where there is a Remainder.}
\begin{nbalignexample}
\nbexample{4331 ÷ 25 &= 173\tfrac{6}{25}}
\end{nbalignexample}

\nbsubparagraph{Rule.}---$5 × 5 = 25$. Divide the dividend ($4331$)
by one of the factors ($5$) and divide the quotient
($866$) by the other factor ($5$). Multiply the \textit{second} %[**TN: original wording "846"]
remainder (in this example it is $1$) by the first
divisor ($5$ or $5 × 1 = 5$) and add the \textit{first} remainder
($1$) to this product (5 or $5 + 1 = 6$); using this
sum ($6$) for the upper fraction and the divisor ($25$)
for the lower part of the fraction ($\frac{6}{25}$) and adding
this fraction to the last quotient ($173$) we get the
answer ($173 \frac{6}{25}$), thus:
\[
\begin{array}{r@{}r@{}r@{}l}
\text{first divisor} \rightarrow 5\;\vline &\multicolumn{2}{@{}r@{}}{\,4331}\\\cline{2-3}
&\llap{5}\;\vline &866&\quad 1\text{ (first remainder)}\\\cline{3-3}
&&173&\quad 1\text{ (second remainder)}
\end{array}
\]
and by the above rule the remainders give $6$ and
by writing this over the $25$ we get the full result:
\[
173 \tfrac{6}{25} \nbanswer
\]

\nbparagraph{(5) To Divide any Number by a Number Formed of Three Factors.}%[**TN: original wording "Any"]
\index{Dividing any number by a number formed of three factors}%
\index{Division, to divide any number by a number formed by three factors}%
\begin{nbalignexample}
\nbexample{703025 ÷ 126 &= 5579\tfrac{71}{126}}
\end{nbalignexample}

\nbsubparagraph{Rule.}---The factors of $126$ are $2$, $7$, and $9$ because
$2 × 7 × 9 = 126$. Divide the dividend by the first
factor ($2$, which in this case gives a quotient of
$351512$ with a remainder of $1$); next divide this
quotient ($351512$) by the second factor ($7$, which
%-----File: 068.png-----
gives in this example a quotient of $50216$ with no
remainder); and finally divide the last quotient
($50216$) by the third factor ($9$, which gives a quotient
of $5579$ with a remainder of $5$), thus:
\[
\begin{array}{lr@{}r@{}r@{}r@{}l}
\text{First divisor}  \rightarrow& 2\;\vline &\multicolumn{3}{r@{}}{703025\;}\\\cline{3-5}
\text{Second divisor} \rightarrow&& \llap{7}\;\vline &\multicolumn{2}{r@{}}{351512}&\quad 1 \leftarrow \text{first remainder}\\\cline{4-5}
\text{Third divisor} \rightarrow&&&\llap{9}\;\vline &50216&\quad 0 \leftarrow \text{second remainder}\\\cline{5-5}
&&&&5579&\quad 5 \leftarrow \text{third remainder}
\end{array}
\]

In order to find the real quotient of the dividend
multiply the third remainder ($5$) by the first two
divisors ($5 × 2 × 7= 70$); multiply the second
remainder ($0$) by the first divisor ($2 × 0 = 0$) and
now add the first remainder ($1$) to the sum of the
two products just obtained, thus:
\[
70 + 0 + 1 = 71
\]
and since $126$ is the divisor the remainder is $\frac{71}{126}$
and hence the full result is
\[
5579 \tfrac{71}{126} \nbanswer
\]

\textbf{To Check the Work.}---The two following methods
for checking results in division are short and
good.
\nbsubparagraph{Check I\@. By Casting out the Nines.}%
\index{Checking division by casting out Nines}%
\index{Division, checking by casting out nines}
\begin{nbalignexample}
\nbexample{&9771 ÷ 232 = 42 \tfrac{27}{232}}\\
&\text{or}&&9771 = 232 + 27 %[**TN: presumably means remainder is 27]
\end{nbalignexample}

By the same process of casting out the nines as
for the multiplication check, see page~\pageref{section:nines}, we have: %[**TN: original page number "39", should have been "53"]
%-----File: 069.png-----
\[
\begin{array}{l}
\text{dividend} \rightarrow 9771 = 6 \text{ which is the excess of nines in } 9771\\
\text{divisor} \rightarrow 232 = 7 \text{ which is the excess of nines in } 232\\
\text{quotient} \rightarrow 42 = 6 \text{ which is the excess of nines in } 42\\
\text{remainder} \rightarrow 27 = 0
\end{array}
\]

Now by multiplying the excess of nines in the
divisor by those in the quotient and adding the
excess of nines in the remainder then---
\[
6 × 7 + 0 = 42
\]
and since the excess of nines in $42$ is $6$ and the
excess of nines in the dividend is $6$ the work is
correct.

\nbsubparagraph{Check II\@. The Lightning Method.}%
\index{Checking division by the lightning method}%
\index{Division, checking by the lightning method}%
\index{Lightning method of checking division}%
---This check
was devised by Virgil D.~Collins and is based on
the method of casting out nines, or the \textit{unitate
method} as explained in the \textit{Lightning Check of
Multiplication}.
\begin{nbalignexample}
\nbexample{9771 ÷ 232 &= 42\tfrac{27}{232}}
\end{nbalignexample}

Now taking the unitates we have---
\[
\begin{array}{r@{}c@{}r@{}l}
9771 = 6& \text{ which is the unitate of } &9771\\
 232 = 7& \text{``\qquad``\qquad``}        &232\\
  42 = 6& \text{``\qquad``\qquad``}        &42 &\text{ and}\\
  27 = 9& \text{``\qquad``\qquad``}        &27
\end{array}
\]

Now since this method is based on the casting
out of nines, whenever we get $9$ as a unitate it is
called a cipher, thus:
\[
27 = 9 = 0
\]

Next multiplying the unitate of the divisor by
%-----File: 070.png-----
the unitate of the quotient and adding the unitate
of the remainder we have---
\[
6 × 7 + 0 = 42
\]
and since the unitate of $42$ is $6$ and the unitate of
$9771$ is $6$, the division is proved correct. All of
these operations should of course be done mentally
in order to do the work with lightninglike rapidity.
%-----File: 071.png-----

\nbchapter{VI}{Short Cuts in Fractions}

\begin{nbchaptercontentsone}
  Aliquot Parts:\\
  \quad Table A\\
  \quad \rlap{\quad``}\phantom{Table} B\\
  \quad \rlap{\quad``}\phantom{Table} C\\
  Decimals:\\
  \quad To Change a Common Fraction into a Decimal\\
  \quad Addition of Decimals\\
  \quad Subtraction of Decimals\\
  \quad Multiplication of Decimals\\
  \quad Figuring Percentage by Decimals\\
  \quad Division of Decimals:\\
  \quad\quad To Divide a Whole Number into a Whole Number when there is a Remainder\\
  \quad\quad To Divide a Decimal into a Whole Number\\
  \quad To Change a Decimal into a Common Fraction
\end{nbchaptercontentsone}

\nbparagraph{Aliquot Parts.}\index{Aliquot parts}%
---In the business world one of
the chief uses of common fractions is in working
out examples in which \textit{aliquot parts} are concerned.

1. An aliquot part is a number which is contained
in another number an exact number of
times, and since our money system is based on
$100$ the aliquot parts of $\$1.00$ are very largely
used.

The aliquot parts of $100$ which consist of whole
numbers are $2$, $4$, $5$, $10$, $20$, $25$, and $50$ and
hence---
%-----File: 072.png-----
{\small
\settowidth{\TmpLen}{Equivalent}
\renewcommand\arraystretch{1.2}
\setlength\arraycolsep{1em}
\[
\begin{array}{ccccc}
&&\textsc{Table A}\\
\parbox[c]{\TmpLen}{\centering Aliquot Part} &&
\parbox[c]{\TmpLen}{\smallskip\centering Equivalent Parts\index{Equivalent parts}\smallskip} &&
\parbox[c]{\TmpLen}{\centering Whole Number}\\
%
\Z2 & \text{is} & \frac{1}{50} & \text{of} & 100\\
\Z4 & \text{``} & \frac{1}{25} & \text{``} & 100\\
\Z5 & \text{``} & \frac{1}{20} & \text{``} & 100\\
 10 & \text{``} & \frac{1}{10} & \text{``} & 100\\
 20 & \text{``} & \frac{1}{5}  & \text{``} & 100\\
 25 & \text{``} & \frac{1}{4}  & \text{``} & 100\\
 50 & \text{``} & \frac{1}{2}  & \text{``} & 100
\end{array}
\]
}

2. Another kind of aliquot parts of $100$ consists
of mixed numbers. It will be seen from the
following table that the numerators of the fractions
representing the \textit{equivalent parts} are in each
case $1$.

{\small
\settowidth{\TmpLen}{Equivalent}
\renewcommand\arraystretch{1.2}
\setlength\arraycolsep{1em}
\[
\begin{array}{ccccc}
&&\textsc{Table B}\\
%
\parbox[c]{\TmpLen}{\centering Aliquot Parts} &&
\parbox[c]{\TmpLen}{\smallskip\centering Equivalent Parts\smallskip} &&
\parbox[c]{\TmpLen}{\centering Whole Number}\\
%
\Z6 \frac{1}{4} & \text{is} & \frac{1}{16} & \text{of} & 100\\
\Z6 \frac{2}{3} & \text{``} & \frac{1}{15} & \text{``} & 100\\
\Z8 \frac{1}{3} & \text{``} & \frac{1}{12} & \text{``} & 100\\
 12 \frac{1}{2} & \text{``} & \frac{1}{8}  & \text{``} & 100\\
 16 \frac{2}{3} & \text{``} & \frac{1}{6}  & \text{``} & 100\\
 33 \frac{1}{3} & \text{``} & \frac{1}{3}  & \text{``} & 100\\
\end{array}
\]
%
\medskip
%
\settowidth{\TmpLen}{Quantity}
\setlength\arraycolsep{0em}
\[
\begin{array}[t]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}cc}
\multicolumn{6}{c}{\text{Examples}}\\
\multicolumn{3}{l}{\parbox[t]{\TmpLen}{\centering Quantity \phantom{Lower}\smallskip}} &
\multicolumn{2}{@{}l}{\parbox[t]{\TmpLen}{Cents}}&
\multicolumn{1}{@{}l}{\parbox[t]{\TmpLen}{Dollars}}\\
&32    & × &  \Z6 \frac{1}{4} &=& \Z2.00\\
&45    & × &  \Z6 \frac{2}{3} &=& \Z3.00\\
&24    & × &  \Z8 \frac{1}{3} &=& \Z2.00\\
&64    & × &   12 \frac{1}{2} &=& \Z8.00\\
&36    & × &   16 \frac{2}{3} %[**TN: original wording "16 1/6"]
&=& \Z6.00\\
&42    & × &   33 \frac{1}{3} &=&  14.00\\
\end{array}
\qquad\qquad\qquad
\settowidth{\TmpLen}{Equivalent Parts}
\begin{array}[t]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}
\multicolumn{5}{c}{\text{Solutions}}\\
\multicolumn{4}{c}{\parbox[t]{\TmpLen}{\centering Equivalent Parts Lower Half\smallskip}}&
\multicolumn{1}{@{}r}{\parbox[t]{2em}{Ans.}}\\
3200 &÷    &16  &=& \$2.00\\
4500 &÷    &15  &=& \Z3.00\\
2400 &÷    &12  &=& \Z2.00\\
6400 &÷    &\Z8 &=& \Z8.00\\
3600 &÷    &\Z6 &=& \Z6.00\\
4200 &÷    &\Z3 &=&  14.00
\end{array}
\]
}
%-----File: 073.png-----

\nbsubparagraph{Rule A.}---To multiply any number by any one
of the aliquot parts given in Table~II, simply
annex two ciphers to the first number and divide
it by the lower part of the fraction of the equivalent
part; thus since the lower part of the equivalent
part of $12 \frac12$ is $8$ the number representing the
quantity $64$ is divided by $8$ after two ciphers have
been annexed to it, which makes it $6400$, thus:
\begin{align*}
64 × 12 \tfrac{1}{2} &= 800\\
6400 ÷ 8  &= 800 %[**TN: original wording "6400 ÷ 6"]
\end{align*}

3. A third class of aliquot parts is where the
upper part of the equivalent fractions are larger
than $1$, as shown in the following table:
{\small
\settowidth{\TmpLen}{Equivalent}
\renewcommand\arraystretch{1.2}
\setlength\arraycolsep{1em}
\[
\begin{array}{ccccc}
&&\textsc{Table C}\\[0.5em]
%
\parbox[c]{\TmpLen}{\centering Aliquot Parts} &&
\parbox[c]{\TmpLen}{\centering Equivalent Parts\smallskip} &&
\parbox[c]{\TmpLen}{\centering Whole Number}\\
13 \frac{1}{3} & \text{is} & \frac{2}{15} & \text{of} & 100\\%[**TN: original wording "= 100"]
53 \frac{1}{3} & \text{``} & \frac{8}{15} & \text{``} & 100\\
62 \frac{1}{2} & \text{``} & \frac{5}{8}  & \text{``} & 100\\
66 \frac{2}{3} & \text{``} & \frac{2}{3}  & \text{``} & 100\\
\end{array}
\]
%
\medskip
%
\settowidth{\TmpLen}{Quantity}
\setlength\arraycolsep{0em}
\[
\begin{array}[t]{@{}c@{\;}c@{\quad}c@{\quad}c@{\quad}cc}
\multicolumn{6}{c}{\text{Examples}}\\
\multicolumn{3}{l}{\parbox[t]{\TmpLen}{\centering  Quantity \phantom{Frac} \phantom{Upper}\smallskip}} &
\multicolumn{2}{@{}l}{\parbox[t]{\TmpLen}{Cents}}&
\multicolumn{1}{@{}l}{\parbox[t]{\TmpLen}{Dollars}}\\
&\Z45 & × &  13 \frac{1}{3}%[**TN: original wording "13 1/2"]
&=& \ZZ600\\
& 450 & × &  53 \frac{1}{3} &=&  24000\\
&\Z40 & × &  62 \frac{1}{2} &=& \ZZ2500\\
&\Z18 & × &  66 \frac{2}{3} &=& \ZZ1200
\end{array}
\qquad\qquad
\settowidth{\TmpLen}{Upper Part}
\begin{array}[t]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}
\multicolumn{9}{@{}c@{}}{\text{Solutions}}\\
\multicolumn{4}{@{}r@{}}{\parbox[t]{\TmpLen}{\centering Equivalent Fraction Lower Part\smallskip}}&
\multicolumn{4}{@{}c@{}}{\parbox[t]{\TmpLen}{\centering Equivalent Fraction Upper Part\smallskip}}&
\multicolumn{1}{@{}r}{\parbox[t]{2em}{Ans.}}\\
\Z4500 & ÷ &  15 & = & \Z300 & × & 2 & = & \Z6.00\\
 45000 & ÷ &  15 & = &  3000 & × & 8 & = &  24.00\\
\Z4000 & ÷ & \Z8 & = & \Z500 & × & 5 & = &  25.00\\
\Z1800 & ÷ & \Z3 & = & \Z600 & × & 2 & = &  12.00
\end{array}
\]
}% end small, etc

\medskip
\nbsubparagraph{Rule B.}---To multiply any mixed number by
an aliquot part of $100$ shown in Table~3 simply add
two ciphers to the number, divide it by the lower
%-----File: 074.png-----
half of the equivalent fraction and multiply the
quotient thus obtained by the upper half of the
equivalent fraction, thus:
\begin{align*}
18 × 66 \tfrac{2}{3} &= 1200\\
1800 ÷ 3             &= 600 × 2 = 1200 \nbanswer
\end{align*}

\nbsection*{DECIMALS}
\index{Decimals}

Very often it is a decided short cut in working
an example to change a common fraction into a
decimal fraction, or \textit{decimal} as it is called for
short.

\nbparagraphdisplay{To Change a Common Fraction into a Decimal.}%
\index{Decimal, to change a common fraction into a}%
\index{Fraction, to change a common fraction into a decimal}
\begin{nbalignexample}
\nbexample{\tfrac{3}{4} = .75 &\text{ or } \tfrac{75}{100}}
\end{nbalignexample}

\nbsubparagraph{General Rule.}---Divide the denominator or lower
part of the fraction into the numerator, or upper
part of the fraction.

Thus to change $\frac34$ to a decimal divide $4$ into $3$
and we get $.75$ as a result, thus:
\[
\begin{array}{r@{}r@{}}
4 \; & \vline \; 3.00 \\\cline{2-2}
     &            .75
\end{array}
\]

In changing common fractions into decimals the
former very often cannot be changed into the
latter and not leave a remainder. But this fact
does not at all prevent decimals from being used
for all ordinary calculations, because in most
examples the remainder is so slight it does not
affect the general result.

In some cases the decimal may run up into the
%-----File: 075.png-----
tens and hundred thousandths; for instance take
$\frac{4}{9}$ and change it into a decimal, then---
\[
\begin{array}{r@{}r@{}}
9 \; & \vline \; 4.0000000\\\cline{2-2}
     &            .4444444
\end{array}
\]
From this it will be seen that the decimal will
never be complete and so for all practical work it is
usual to call it simply $.44$.

If the result is a decimal such as $.8359$ it is the
common practice to call it $.84$ because the last
figure, $9$, is nearer $10$ than it is to $1$ and so we call it
$10$; and if we call it $10$ then the next figure, which
is $5$, would be called a $6$, thus $.836$. Now $6$ is nearer
$10$ than it is to $1$, so that calling the $6$ a $10$ the $3$
becomes $4$, hence we write $.84$, and this is close
enough for all ordinary purposes.

\nbparagraph{Addition of Decimals.}%
\index{Addition of decimals}%
\index{Decimals, addition of}%
---The operation of adding
decimals is exactly the same as that of adding
whole numbers, thus:
\[
\begin{array}{r@{\;}|@{\,}ll}
 .& 735&\\
 .& 720&\\
 .& 012&\\
 \cline{2-2}
1.& 467&\text{ or } 1.47 \nbanswer %[**TN: original wording ".467 or .47"]
\end{array}
\]

Before starting to add put the decimal points
down for the answer in the column of decimal
points as shown above and then it will always be in
the right place.

\nbparagraph{Subtraction of Decimals.}---In the subtraction of
decimals set the subtrahend and minuend down
exactly as in the operation of subtracting whole
%-----File: 076.png-----
numbers and place all the decimal points in a column
as described above under the caption of \textit{Addition
of Decimals}, thus:
\[
\begin{array}{r@{}l}
2.821&\\
 .960&\\
\cline{1-1}
1.861&\text{ or } 1.86 \nbanswer
\end{array}
\]

\nbparagraph{Multiplication of Decimals.}%
\index{Decimals, multiplication of}%
\index{Multiplication of decimals}%
---While the operation
of multiplying decimals is exactly the same
as that of multiplying whole numbers care must
be taken to get the decimal point in the right
place.

The rule for placing the decimal point in the
product obtained by multiplying decimals is this:

\nbsubparagraph{Rule.}---Add the number of decimal places in
the multiplicand to those in the multiplier and,
beginning at the right of the product, point off the
sum obtained, thus:
\[
\begin{array}{@{}r@{}r@{}ll}
  &4.325&&\text{(Three places in the multiplicand) }+\\
  &  .25&&\text{(Two places in the multiplier) }=\\
\cline{2-2}
  &21625&&\\
  & 8650\Z&&\\
\cline{1-2}
1.&08125 &&\text{(Five places in the product)}\\
(.&54321&)&
\end{array}
\]

{\nbstretchyspace
\nbparagraph{Figuring Percentage by Decimals.}%
\index{Decimals, figuring percentage by}%
\index{Percentage by decimals, figuring}%
---Figuring
money percentages by means of decimals is short,
quick, and easy for it is only necessary to multiply
the per cent by the amount.}

Per cent is always taken as hundredth parts of
a dollar, for instance, instead of writing 5 per cent
%-----File: 077.png-----
when we are doing a problem of this sort we simply
write $.05$ or five one-hundredths ($\frac{5}{100}$)

\nbsubparagraph{Example}: Suppose that you have a first mortgage
of $\$97.00$ on your new bungalow at $5$ per cent
per annum and you want to know how much you
will owe at the end of the first year. Simply
multiply $\$97.00$ by $5$, thus:
\[
\arraycolsep=0.1em
\begin{array}{rl}
\$97.00&\\
    .05&\\
\cline{1-1}
\$4.8500&\text{ or } \$4.85 \nbanswer
\end{array}
\]

Point off four places, because there are two
places in the multiplicand ($00$) and two places in
the multiplier ($05$) pointed off and together these
make four places.

\nbparagraph{Division of Decimals.}%
\index{Decimals, division of}%
\index{Division of decimals}%
---The division of decimals
is perhaps a little harder than the foregoing
operations; however there are very few examples
in division, except those which come out even,
which do not employ decimals.

\nbparagraphdisplay{To Divide a Whole Number into a Whole Number
when there is a Remainder.}
\index{Decimals, to divide a whole number into a whole number when there is a remainder}
\index{Division of decimals, to divide a whole number into a whole number when there is a remainder} %[**TN: original wording "no remainder"]

\nbsubparagraph{Rule A}.---Divide as usual, except a decimal
point should be placed after the last figure of the
dividend and two or more ciphers must be annexed
to it, according to the number of decimal places
you want to have in the answer, thus:
\begin{nbalignexample}
\nbexample{143 ÷ 35 &= 4.085 \nbanswer}
\end{nbalignexample}

Now by annexing three ciphers, $143$ with the
%-----File: 078.png-----
decimal point becomes $143.000$, so we write the
example like this:
\[
\arraycolsep=0em
\begin{array}{r@{\nbdivspace}c@{\nbdivspace}llllll}
  & &\multicolumn{4}{r}{4.085}   &&\nbanswer\\
\cline{2-6}
35&\big)&\multicolumn{4}{l}{143.000} &&\\
  & &14 &0&&&&\\
\cline{3-5}
  & &     &\multicolumn{2}{l}{300}&&&\\
  & &     &\multicolumn{2}{l}{280}&&&\\
\cline{4-7}
  & &     &\multicolumn{4}{l}{\Z200}&\\ %[**TN: adjusted alignment]
\end{array}
\]

It will be seen that the decimal point in the
answer is placed directly above that in the dividend,
the process of dividing being exactly the same as
for ordinary division.

\nbparagraphdisplay{To Divide a Decimal into a Whole Number.}
\index{Decimals, to divide a decimal into a whole number}
\index{Division, to divide a decimal into a whole number}

\nbsubparagraph{Rule B}.---Annex as many ciphers to the dividend
as there are decimal places in the divisor
and this changes the problem into one like that
described in Rule~\textit{A}, that is, one where a whole
number is divided into a whole number thus:
\begin{nbalignexample}
\nbexample{625 ÷ .25 &= 2500}
\end{nbalignexample}

In this example $625$ with two ciphers added
becomes $62500$ because there are two decimal
places in the decimal $.25$. The example for this
reason is written---
\[
\arraycolsep=0em
\begin{array}{r@{\nbdivspace}c@{\nbdivspace}lllll}
  & & \multicolumn{4}{r}{2500.}&\nbanswer\\
\cline{2-6}
25&\big)&\multicolumn{4}{l}{62500.}&\\
  & &                   50&&&&\\
\cline{3-4}
  & &                   12&5&&&\\
  & &                   12&5&&&\\
\cline{3-5}
  & &                     &0&0&&
\end{array}
\]
%-----File: 079.png-----

Here the annexing of two ciphers to the dividend
changes the decimal into a whole number. In
other words we have multiplied both the divisor and
the dividend by 100 by means of a short mental
cut, thus:
\begin{align*}
625 × 100 &= 62500\\
.25 × 100 &= 25
\end{align*}

So that we have changed the divisor which was
a decimal into a whole number.

\nbparagraphdisplay{To Divide a Decimal by a Decimal.}
\index{Decimals, to divide a decimal into a decimal}
\index{Division, to divide a decimal by a decimal}

\nbsubparagraph{Rule C.}---To divide a decimal by a decimal
start at the left of the decimal point of the dividend
and point off as many places as there are
places in the divisor, thus:
\begin{nbalignexample}
\nbexample{1.548 ÷ .36 &= 4.3 \nbanswer}
\end{nbalignexample}

Since there are two places in the divisor $1.548$
becomes $154.8$ and the operation is---
\[
\arraycolsep=0em
\begin{array}{r@{\nbdivspace}c@{\nbdivspace}llllll}
  & &\multicolumn{4}{l}{\ZZ4.3}   &&\nbanswer\\
\cline{2-6}
36&\big)&\multicolumn{4}{l}{154.8} &&\\
  & &14 &4&&&&\\
\cline{3-5}
  & &     &\multicolumn{2}{l}{108}&&&\\
  & &     &\multicolumn{2}{l}{108}&&&\\
\cline{4-7}
\end{array}
\]

\nbparagraph{To Change a Decimal into a Common Fraction.}%
\index{Decimals, to change a decimal into a common fraction}%
\index{Fraction, to change a decimal into a common fraction}%
---Occasionally
it is necessary to change a decimal
into a common fraction, as for

\nbsubparagraph{Example:}\quad Change $.725$ into a common fraction.

\nbsubparagraph{Rule.}---First write the decimal as if it were a
whole number, that is without the decimal point.
%-----File: 080.png-----
Then draw the fraction line, count the number of
decimal places in the decimal and set down an
equal number of ciphers and prefix a 1, thus:
\[
\frac{\Z725}{1000}
\]
Since there are three decimal places in the
decimal, in the above example there must be three
ciphers in the denominator or the lower part of
the fraction.
%-----File: 081.png-----

\nbchapter{VII}{Extracting Square and Cube Roots}
\index{Extracting square and cube root}

To figure out the square or cube root of a number
is not an altogether simple operation, but if you
set your work down systematically and follow the
rules here given you will have no trouble.

\nbsection*{SQUARE ROOT}
\index{Square root, extracting}

To begin with we know that $1 = 1^2$, $100 = 10^2$,
$10000 = 100^2$, etc., and this being true then the
square root of any number between $1$ and $100$ must
be some number between $1$ and $10$.

The above statement may be put in another
form by saying that the square root of any two-figure
number is a one-figure number, the square
root of any three or four-figure number is a two-figure
number, and so on.

Now if we have a large number which we want
to find the square root of, say for example
\[
4225
\]
the first step is to divide it into two numbers of
two figures each, thus:
%-----File: 082.png-----
\[
\begin{array}{cc}
\text{2d pair} & \text{1st pair}\\
\overbrace{\quad}   & \overbrace{\quad}\\
42             & 25
\end{array}
\]
where $25$ may be considered as the first number
pair and $42$ as the second number pair and hence
our answer must be a two-figure number.

To find the square root of a four-figure number
such as our example calls for, work it out in the
following manner:
\[
\arraycolsep=0em
\begin{array}{cccccccccccc}
\multicolumn{3}{c}{\text{2d pair}} &\qquad&\qquad&
\multicolumn{3}{c}{\text{1st pair}}&\qquad&\qquad&
\multicolumn{2}{c}{\text{square root}}\\
 & \overbrace{\quad}&\ZZZ&&&\ZZZ&\overbrace{\quad}&\ZZZ\\
 & 42 &&&&& 25 &&&&\vline\;65&\nbanswer\\\cline{11-11}
 & 36 &&&&& \\\cline{2-2}
\multirow{2}{*}{125\:} & \vline\hfill \Z6 \hfill&&&&& 25\\
 & \vline\hfill \Z6 \hfill&&&&& 25\\\cline{2-7}
\end{array}
\]

\nbsubparagraph{First step.}---Take the nearest square of the
second pair, which in this case is the square of $6$.
Set the $6$ down in the answer and set the square
($36$) down under the second pair, then subtract
and bring down the first pair.

\nbsubparagraph{Second step.}---Next multiply the $6$ set down in
the answer by $2$; this gives $12$ as the product and
this ($12$) is called the \textit{trial divisor}. Divide the $12$
into the first two figures of the remainder ($625$),
or in other words divide the $12$ into the $62$ and as
it is easy to see that it will go $5$ times set down $5$
in the answer.

\nbsubparagraph{Third step.}---Now annex a $5$ to the $12$, making
$125$, and multiply this number by the $5$ in the
answer. In this case there is no remainder and so
the square root of $4225$ is $65$. \nbanswer
%-----File: 083.png-----

To extract the square root of a larger number,
say
\[
43681
\]
proceed as follows:

\[
\arraycolsep=0em
\begin{array}{rcc@{\qquad}c@{\qquad}cc}
\multicolumn{2}{c}{\text{3d pair}}\qquad   & \text{2d pair} & \text{1st pair} & \multicolumn{2}{c}{\text{square root}}  \\
\multicolumn{2}{c}{\overbrace{\quad}}\qquad&\overbrace{\quad}&\overbrace{\quad}&  \\
            \ZZ4\;&                    &         36      &       81        &\vline \hfill\;209&\nbanswer\\\cline{5-5}
            \ZZ4\;&                    &                 &                 &  \\\cline{1-1}
       409\;\vline&                    &         36      &       81        &  \\
            \vline&                    &         36      &       81        &  \\\cline{2-4}
\end{array}
\]

\nbsubparagraph{Solution.}---Since $4$ is the trial divisor and is
not contained in $3$, set down a cipher in the answer
and annex another cipher to the $4$, which makes
the $4$ a $40$.

Then bring down the first number pair ($81$);
using $40$ as a trial divisor divide $40$ into $368$, this
gives $9$ as the quotient and hence annex $9$ to the
$40$ and set down $9$ in the answer also. The final
step is to multiply $409$ by the $9$ in the answer.

When extracting the square root of a number
which does not come out even, that is a number
whose square root is a whole number and a decimal,
proceed as usual but annex a pair of ciphers to the
number and extract the square root to as many
decimal places as you wish.

\nbsection*{CUBE ROOT}
\index{Cube root, extracting}

To extract the cube root of a number is not a
harder operation than finding the square root.

Since we know that $1 = 1^3$, $1,000 = 10^3$, and
%-----File: 084.png-----
$1,000,000 = 100^3$, etc., it must be clear that the
cube root of any number between $1$ and $1,000$
must be some number which lies between $1$ and
$10$; in a like manner the cube root of any number
between $1,000$ and $1,000,000$ must be some number
between $10$ and $100$.

Expressing this in another way we may say
that the cube root of any three-figure number is a
one-figure number while that of any five-figure
number is a two-figure number and so forth.

If then we have a large number and we want to
take its cube root the first step is to divide it into
parts of three figures starting from the right; take
for example:
\[
91,125
\]
divide it into parts of three figures, thus:
\[
\begin{array}{c@{\qquad\qquad}c}
\text{1st part} & \text{2d part}\\
\overbrace{\quad}   & \overbrace{\quad}\\
91             & 125
\end{array}
\]

Now to find the cube root of 91,125 proceed as
follows:
\begin{gather*}
\arraycolsep=0em
\begin{array}{cccc@{\qquad\quad}cc}
\multicolumn{3}{c}{\text{1st part}} & \text{2d part} & \multicolumn{2}{c}{\text{cube root}}  \\
\multicolumn{3}{c}{\overbrace{\quad}}&\overbrace{\quad} \\
        \qquad\qquad&91&\qquad\qquad &                        125     &\vline \hfill\;45&\nbanswer\\\cline{5-5}
                    &64&&         \\\cline{2-2}
                    &27&\\
\end{array}\\
\nbsubparagraph{Steps in the Solution}\\
\begin{aligned}
    3 × 40^2 =  4800&\\
3 × (40 × 5) = \Z600&\\
         5^2 = \ZZ25&\\
  \text{Adding\quad} \nboverline{5425}&\\
\end{aligned}
\end{gather*}
%-----File: 085.png-----

\nbsubparagraph{First step.}---Take the nearest cube of the first
number part ($91$). In this case it is 64 and the
cube root of $64$ is $4$ since $4 × 4 × 4 = 64$. Set
this cube root ($4$) down in the answer and then
subtract the $64$ from the first part number ($91$).

\nbsubparagraph{Second step.}---The remainder is $27,125$ and in
order to find a number that will go into $27,125$
evenly (if the number you are taking the cube
root of is a perfect cube) proceed as follows:

\nblabel{a} Annex a cipher to the $4$ in the answer,
square it and multiply it by $3$ thus:
\[
40^2 × 3 = 4800
\]

\nblabel{b} The first two figures of this product, that is,
$48$, is what is known as the trial divisor. Divide
the first three figures of $27125$ by $48$, thus:
\[
271 ÷ 48 = 5
\]

\nblabel{c} Next annex a cipher to the $4$ in the answer,
multiply it by the $5$ just found, and then by $3$
thus:
\[
3 × (40 × 5) = 600
\]

\nblabel{d} Now square the $5$ just found and add the
three products together thus:
\[
4800 + 600 + 25 = 5425
\]

\nblabel{e} This number, $5425$, when multiplied by the
$5$ found by the trial divisor in part \textit{b} should equal
the remainder found in step \textit{a}, that is $5425 × 5 =
27125$.
%-----File: 086.png-----

\nblabel{f} So that the $5$ should now be set down in the
answer and the cube root of $91,125$ is $45$.

Should the number not be a perfect cube and
the root does not come out even then annex three
ciphers to the number in parts of three, when the
root may be carried out to as many decimal places
as desired.
%-----File: 087.png-----

\nbchapter{VIII}{Useful Tables and Formulas}
\index{Formulas, useful}
\index{Tables and formulas, useful}
\index{Useful tables and formulas}

\begin{nbchaptercontentsnum}{3.5em}
\item[I.] The Roman Numerals
\item[II.] Measure of Length
\item[III.] Other Measures of Length
\item[IV.] Measure of Area
\item[V.] Land Measure
\item[VI.] Measure of Volume
\item[VII.] Measure of Capacity (Liquid)
\item[VIII.] Apothecaries' Fluid Measure
\item[IX.] Dry Measure
\item[X.] Avoirdupois Weight
\item[XI.] Apothecaries' Weight
\item[XII.] Troy Weight
\item[] \textit{Weight by Carat}
\item[] \textit{Fineness of Gold}
\item[XIII.] Comparison of Weights
\item[XIV.] Household Measure
\item[] \textit{Rule for Changing Parts by Volume to Parts by Weight} %[**TN: contents order changed to match main text]
\item[XV.] Miscellaneous Measure
\item[] \textit{Measures of Circles and Angles}
\item[XVI.] Circular or Angular Measure by Degrees
\item[XVII.] Angular Measure by Radians
\item[XVIII.] Astronomical Measure
\item[XIX.] Geographical Measure
\item[XX.] Decimal Equivalents of Fractions of an Inch
\item[]\quad\textbf{Metric System of Weights and Measures}
\item[XXI.] Measure of Length
\item[XXII.] Measure of Surface
%-----File: 088.png-----
\item[XXIII.] Measure of Land
\item[XXIV.] Measure of Volume (Cubic Measure)
\item[XXV.] Measure of Volume (Wood Measure)
\item[XXVI.] Measure of Volume (for Liquids and Solids)
\item[XXVII.] Measure of Weight
\item[] \textit{A Key to the Metric System}
\item[] \textit{Rules for Measuring Surfaces and Solids}\\
\begin{tabular}{l@{\hspace{8em}}l}
A---Triangle  & Square or Rhombus\\
B---Triangle  & Trapezium\\
Parallelogram & Circle\\
\multicolumn{2}{c}{Ellipse}
\end{tabular}
\item[] \textit{Rules for Measuring Circles and Spheres}\\
\begin{tabular}{l@{\hspace{3em}}l}
Circumference of a Circle&   Surface of a Sphere\\
Area of a Circle         &     Volume of a Sphere
\end{tabular}
\item[] \textit{Rules for Computing Simple Interest}
\end{nbchaptercontentsnum}

\nbtabletitle{Table I, The Roman Numerals}
\index{Numerals, Roman}
\index{Roman numerals}
\nbreflabel{table:I}

\begin{center}
\begin{tabular}{c@{\qquad\qquad\qquad}c}
\small
  \begin{tabular}[t]{r@{\;}c@{\;}l}
      1 & = & I\\
      2 & = & II\\
      3 & = & III\\
      4 & = & IV\\
      5 & = & V\\
      6 & = & VI\\
      7 & = & VII\\
      8 & = & VIII\\
      9 & = & IX\\
     10 & = & X\\
     20 & = & XX\\
     30 & = & XXX\\
     40 & = & XL\\
     50 & = & L\\
     60 & = & LX\\
     70 & = & LXX\\
     80 & = & LXXX\\
     90 & = & XC\\
    100 & = & C
  \end{tabular}&
  \begin{tabular}[t]{r@{\;}c@{\;}l}
          500 & = & D or L\nbrotC\\
        1,000 & = & M or C\nbrotC\\
        2,000 & = & MM or II\nbrotC\nbrotC\nbrotC\\
        5,000 & = & $\overline{\text{V}}$ or L\nbrotC\nbrotC\\
        6,000 & = & $\overline{\text{VI}}$ or LX\nbrotC\nbrotC\\ %[**TN: original wording "MMM", or 3,000]
       10,000 & = & $\overline{\text{X}}$ or C\nbrotC\nbrotC\\
       50,000 & = & $\overline{\text{L}}$ or L\nbrotC\nbrotC\nbrotC\\
       60,000 & = & $\overline{\text{LX}}$ or LX\nbrotC\nbrotC\nbrotC\\ %[**TN: original wording "MMM\nbrotC" or 30,000]
      100,000 & = & $\overline{\text{C}}$ or C\nbrotC\nbrotC\nbrotC\\
    1,000,000 & = & $\overline{\text{M}}$ or C\nbrotC\nbrotC\nbrotC\nbrotC\\
%[**TN: on the following line, original wording omitted " or MM" - this seems the most likely original intention]
    2,000,000 & = & $\overline{\text{MM}}$ or MM\nbrotC\nbrotC\nbrotC\\[1ex]
    \multicolumn{3}{c}{\parbox{14em}{When a line is drawn over a number it means that its value is increased 1000 times.}}
  \end{tabular}
\end{tabular}
\end{center}

%-----File: 089.png-----

\nbsubsection*{English System of Weights and Measures}
\index{Weights and measures, English}
\index{English system, of weights and measures}

\nbtabletitle{Table II, Measure of Length}
\index{Length, English measure of}
\index{Measure of length, English}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
12 inches         &=& 1 foot\\
3 feet            &=& 1 yard\\
$5 \frac12$ yards &=& 1 rod\\
320 rods          &=& 1 statute mile
\end{tabular}
\end{center}

\nbtabletitle{Table III, Other Measures of Length}
\nbreflabel{table:III}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}p{24em}}
1 size    &=& $\frac13$ inch (Used by shoemakers)\\
1 hand    &=& 4 inches (Used in measuring the\ height of horses)\\
1 fathom  &=& 6 feet (Used in measuring the depth of the sea)\\
1 chain   &=& 100 links = 22 yards (Used by surveyors)\\
1 furlong &=& 40 rods = 220 yards\\
1 knot    &=& $1.52 \frac23$ statute miles or 6,086 feet
(Used in measuring distances at sea)\\ %[**TN: missing open bracket added]
\end{tabular}
\end{center}

\nbtabletitle{Table IV, Measure of Area}
\index{Measure of area}
\index{Area, English measure of}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}p{16em}}
\Z144 square inches         &=& 1 square foot\\
\ZZZ9 square feet           &=& 1 square yard\\
\ZZ30$\frac14$ square yards &=& 1 square rod\\
\ZZ16 square rods           &=& 1 square chain\\
\ZZ10 square chains         &=& 1 acre\\
\Z160 square rods           &=& 1 acre\\
 4840 square yards          &=& 1 acre\\
\Z640 acres                 &=& 1 square mile\\
\Z100 square feet           &=& 1 square (Used in measuring
roofing, flooring, etc.)\\
\end{tabular}
\end{center}
%-----File: 090.png-----

\nbtabletitle{Table V, Land Measure}
\index{Land measure, English}
\index{Measure, English land}

{\setlength\parskip{2pt plus 1pt} %to avoid \frac12 on consecutive lines clashing
A township has $36$ sections, each $1$ mile square.

A section is $640$ acres.

A quarter section $\frac12$ mile square is $160$ acres.

An eighth section $\frac12$ mile long and $\frac14$ mile wide is
80 acres.

Sections are all numbered from $1$ to $36$, commencing
at the northeast corner.

One acre contains $4,840$ square yards, or $43,560$
square feet.

A lot $208 \frac23$ feet square contains 1 acre.
}

\nbtabletitle{Table VI, Measure of Volume}
\index{Measure of volume, English}
\index{Volume, English measure of}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
1728 cubic inches  &=& 1 cubic foot\\
24$\frac34$ cubic feet  &=& 1 perch\\
27 cubic feet      &=& 1 cubic yard\\
128 cubic feet     &=& 1 cord  (Used in measuring
wood, etc.)\\
1 cubic yard       &=& 1 load  (Used in measuring
earth, etc.)
\end{tabular}
\end{center}

\nbtabletitle{Table VII, Measure of Capacity---Liquid}
\index{Capacity, English measure of liquid}
\index{Liquid measure of capacity, English}
\index{Measure of liquid capacity, English}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
4 gills              &=& 1 pint\\
2 pints              &=& 1 quart\\
4 quarts             &=& 1 gallon\\
1 gallon             &=& 4 quarts = 8 pints = 32 gills\\
31$\frac12$ gallons &=& 1 barrel\\
63 gallons           &=& 1 hogshead
\end{tabular}
\end{center}
%-----File: 091.png-----

\nbtabletitle{Table VIII, Apothecaries' Fluid Measure}
\index{Apothecaries' fluid measure, English}
\index{Fluid measure, apothecaries', English}
\index{Measure, apothecaries' fluid, English}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
60 minims       &=& 1 fluid drachm
  (f\,\raisebox{2pt}{\large\yogh})\\ %[**TN: \yogh is close in appearance to apothecaries' symbol for drachm]
8 fluid drachms &=& 1 fluid ounce
  (f\,\raisebox{-0.17ex}{\includegraphics[width=0.46em]{images/oz.pdf}})\\ %[**TN: image created by modifying \yogh, for similar style]
16 fluid ounces &=& 1 pint
  (\raisebox{1pt}{$\scriptstyle\bigcirc$})\\
8 pints         &=& 1 gallon (Cong.)
\end{tabular}
\end{center}

\nbtabletitle{Table IX, Dry Measure}
\index{Measure, English dry}
\index{Dry measure, English}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
2 pints  &=& 1 quart\\
8 quarts &=& 1 peck\\
4 pecks  &=& 1 bushel
\end{tabular}
\end{center}

\medskip
\begin{center}\textsc{Measures of Weight}\end{center}
\index{Weight, English measure of}
\index{Measure of weight, English}

Weight is the measure of the attraction between
the earth and bodies upon or near its surface.
There are four kinds of weight used in the United
States and these are (\textit{a}) \textit{avoirdupois}, (\textit{b}) \textit{apothecaries'},
(\textit{c}) \textit{troy}, and (\textit{d}) \textit{metric}.

(\textit{a}) Avoirdupois is the kind of weight that is
used for all ordinary commercial purposes such
as weighing sugar and other staple articles.

(\textit{b}) Apothecaries' weight is the kind used by
druggists in compounding prescriptions; but druggists
buy their drugs and chemicals by avoirdupois
weight.

(\textit{c}) Troy weight is used by goldsmiths and jewellers
and also by the U.S. Mint for weighing coins.

(\textit{d}) Metric weight is used for nearly all scientific
measurement. It will be fully described under
the caption of Metric Weights and Measures.
%-----File: 092.png-----

\nbtabletitle{Table X, Avoirdupois Weight}
\index{Avoirdupois weight, English}
\index{Weight, English avoirdupois}

\begin{center}
\index{Ton, English long}
\index{Ton, English short}
\index{Ton, metric}
\begin{tabular}{l@{\;}c@{\;}l}
\Z16 ounces         &=& 1 pound\\
100 pounds          &=& 1 hundred-weight\\
\Z20 hundred-weight &=& 1 ton = 2000 pounds\\
\multicolumn{3}{l}{A \textit{short} ton is 2000 pounds.}\\
\multicolumn{3}{l}{A \textit{long} ton is 2240 pounds (also called a gross ton).}\\
\multicolumn{3}{l}{A \textit{metric} ton is 2204.6 pounds. (See metric system).}
\end{tabular}
\end{center}

\nbtabletitle{Table XI, Apothecaries' Weight}
\index{Apothecaries' weight, English}
\index{Weight, apothecaries' English}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
 20 grains   &=& 1 scruple\\
\Z3 scruples &=& 1 dram\\
\Z8 drams    &=& 1 ounce\\
 12 ounces   &=& 1 pound
\end{tabular}
\end{center}

\nbtabletitle{Table XII, Troy Weight}
\index{Weight, English Troy}
\index{Troy weight, English}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
24 grains          &=& 1 pennyweight\\
20 pennyweights    &=& 1 ounce\\
12 ounces          &=& 1 pound
\end{tabular}
\end{center}

\nbtabletitle{Weight by Carat}
\index{Carat, weight by}
\index{Weight by carat}
\index{Karat|see{Carat}}

A \textit{carat}, or as it is sometimes written, \textit{karat}, is
the unit used for weighing precious stones. The
international carat is equal to 3.168 grains, or
205 milligrams.

\nbtabletitle{Fineness of Gold}
\index{Fineness of gold}
\index{Gold, fineness of}

The word \textit{carat}, or \textit{karat}, is also used to indicate
the proportion of gold in a gold alloy. The carat
%-----File: 093.png-----
is the 24th part by weight of a gold alloy and is
derived from a weight equal to the $\frac{1}{24}$ part of a
gold mark which was a unit of weight used in many
European countries before the metric system was
adopted. This gold which is said to be 18 carats
fine is an alloy containing $\frac{18}{24}$ or $\frac{3}{4}$ parts of pure
gold, the other $\frac{6}{24}$ being some inferior metal.

\nbtabletitle{Table XIII, Comparison of Weights}
\index{Comparison of English weights}
\index{Weights, comparison of English}

\begin{center}
\begin{tabular}{p{8em}l@{\qquad}l@{\qquad}c}
\multicolumn{1}{c}{\textit{Kind}}&\multicolumn{1}{c@{\qquad}}{\textit{Pound}}
 &\multicolumn{1}{c@{\qquad}}{\textit{Ounce}}&\multicolumn{1}{c}{\textit{Grain}}\\
Avoirdupois\dotfill   &  7000 gr. & $437\frac{1}{2}$           gr. & 1\\
Apothecaries'\dotfill &  5760 gr. & 480\phantom{$\frac{1}{2}$} gr. & 1\\
Troy\dotfill          &  5760 gr. & 480\phantom{$\frac{1}{2}$} gr. & 1
\end{tabular}
\end{center}

\nbtabletitle{Table XIV, Household Measure}
\index{Household measure, English}
\index{Measure, household}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
 120 drops of water               &=& 1 teaspoonful\\
\Z60 drops of thick liquid        &=& 1 teaspoonful\\
\ZZ1 teaspoonful of thick  liquid &=& 1 ounce\\
\ZZ2 teaspoonsful                 &=& 1 dessert-spoonful\\
\ZZ3 teaspoonsful                 &=& 1 tablespoonful\\
\Z16 tablespoonsful               &=& 1 cup\\
\ZZ1 cup                          &=& $\frac{1}{2}$ pint\\
\ZZ1 cup of water                 &=& $\frac{1}{2}$ pound
\end{tabular}
\end{center}

\nbtabletitle{Rule for Changing Parts by Volume or Measure to Parts by Weight}
\index{Measure, rule of changing parts by measure to parts by weight}
\index{Volume, rule for changing parts by volume to parts by weight}
\index{Weight, rule for changing parts by volume or measure to parts by}
\index{Rule for changing parts by volume or measure to parts by weight}

The different substances which are used for
recipes are sometimes given by volume or measure
and sometimes by weight. It is often convenient
%-----File: 094.png-----
to change the parts by volume into parts by weight
and this can be done by multiplying the parts by
volume by the \textit{density}, that is, the \textit{specific gravity},
of the substance which it is desired to change and
the product obtained will be the parts by weight.

\nbtabletitle{Table XV, Miscellaneous Measure}
\index{Measure, miscellaneous}
\index{Miscellaneous measure}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
 12 articles         &=& 1 dozen\\
 12 dozen            &=& 1 gross\\
 24 sheets of paper  &=& 1 quire\\
 20 quires           &=& 1 ream\\
\Z2 reams            &=& 1 bundle\\
\Z5 bundles          &=& 1 bale\
\end{tabular}
\end{center}

\nbtabletitle{Measure of Circles and Angles}
\index{Measure of angles and circles}

Since circles are of varying diameters they are
measured by unit spaces called \textit{degrees}. A degree
($°$) is the 360th part of a circle. Each degree is
further divided into minutes ($'$) and the minutes
into seconds ($''$), as the following table shows.

\textit{Angles} are also measured by degrees; a right
angle or quadrant is $90°$ and hence there are four
right angles or quadrants in a circle.

\nbtabletitle{Table XVI, Circular or Angular Measure by Degrees} %[**TN: original wording "Circular of Angular"]
\index{Angles, measure of}
\index{Angular measure by degrees}
\index{Circles, measure of}
\index{Circular measure, by degrees}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
\Z60 seconds &=& 1 minute\\
\Z60 minutes &=& 1 degree\\
\Z90 degrees &=& 1 quadrant or right angle\\
 360 degrees &=& 1 circle
\end{tabular}
\end{center}
%-----File: 095.png-----

\nbtabletitle{Angular Measure by Radians}
\index{Angular measure by radians}
\index{Measure of angles by radius}
\index{Radians, angular measure by}

Another way to measure an angle is by its
\textit{radian}. A radian is the arc of a circle whose length
is equal to the radius of the circle of which it is a
part. The kind of angles most often used are
right angles and angles less than $90°$.

\nbtabletitle{Table XVII}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
\Z1 minute  &=& 60 seconds\\
\Z1 degree  &=& 60 minutes\\
 90 degrees &=& 1 right angle\\
\Z1 radian  &=& $180 ÷ \pi = 57°\: 17'\: 45''$\\
\Z1 degree  &=& $\pi ÷ 180 = 0.017453$ radian
\end{tabular}\\[1ex]
\raggedright\textit{Note}---For the value of $\pi$ (Greek letter pi) see
Table.
\end{center}

\nbtabletitle{Table XVIII, Astronomical Measure}
\index{Astronomical measure}
\index{Measure, astronomical}

Sometimes astronomers use the following divisions
of the circle:
\begin{center}
\begin{tabular}{l@{\;}l@{\;}c@{\;}c@{\;}l}
  12 & signs of 30 degrees      &each&=& 1 circle\\
 \Z6 & sextants of 60  \hfill''\ \;& '' &=& 1 circle\\
 \Z4 & quadrants of 90 \hfill''\ \;& '' &=& 1 circle
\end{tabular}
\end{center}

\nbtabletitle{Table XIX, Geographical Measure}
\index{Geographical measure}
\index{Measure, geographical}

\begin{center}
\begin{tabular}{l@{\;}c@{\:}p{0.6\linewidth}}
\ZZ1 geographical mile  &=& 6087.15 feet\\
\ZZ1 geographical mile  &=& 1.15287 statute mile\\
\Z60 geographical miles &=& 1 degree of longitude at the equator\\
 360 degrees            &=& the circumference of the  earth at the equator
\end{tabular}
\end{center}
%-----File: 096.png-----

\nbtabletitle{Table XX, Decimal Equivalents of Fractions of an Inch}
\index{Decimal equivalents of fractions of an inch}
\index{Equivalent fractions of an inch, decimal}
\index{Fractions of an inch, decimal equivalents of}

\settowidth{\TmpLen}{\scriptsize Equivalents}
\begin{center}
\footnotesize
\begin{tabular}{c|c|c|c|l||c|c|c|c|l}
\hline\hline
\scriptsize  8ths &
\scriptsize 16ths &
\scriptsize \,32ds\, &
\scriptsize 64ths &
\parbox[c]{\TmpLen}{\scriptsize \centering \vspace{1ex}Decimal Equivalents\vspace{1ex}}&
\scriptsize  8ths &
\scriptsize 16ths &
\scriptsize \,32ds\,  &
\scriptsize 64ths &
\parbox[c]{\TmpLen}{\scriptsize \centering Decimal Equivalents}\\
\hline
\rule{0pt}{3.5ex}  &    &     & \Z1 & \Z.015625 &                    &     &    & 33 & \Z.515625\\
                   &    & \Z1 &     & \Z.03125  &                    &     & 17 &    & \Z.53125\\
                   &    &     & \Z3 & \Z.046875 &                    &     &    & 35 & \Z.546875\\
                   &  1 &     &     & \Z.0625   &                    & \Z9 &    &    & \Z.5625\\
                   &    &     & \Z5 & \Z.078125 &                    &     &    & 37 & \Z.578125\\
                   &    & \Z3 &     & \Z.09375  &                    &     & 19 &    & \Z.59357\\
                   &    &     & \Z7 & \Z.109375 &                    &     &    & 39 & \Z.609375\\
$1\phantom{(\frac{1}{8}'')}$ & & &  & \Z.125    & $5\phantom{(\frac{5}{8}'')}$ & & & & \Z.625\\
                   &    &     & \Z9 & \Z.140625 &                    &     &    & 41 & \Z.640625\\
                   &    & \Z5 &     & \Z.15625  &                    &     & 21 &    & \Z.65625\\
                   &    &     &  11 & \Z.171875 &                    &     &    & 43 & \Z.671875\\
                   &  3 &     &     & \Z.1875   &                    &  11 &    &    & \Z.6875\\
                   &    &     &  13 & \Z.203125 &                    &     &    & 45 & \Z.703125\\
                   &    & \Z7 &     & \Z.21875  &                    &     & 23 &    & \Z.71875\\
                   &    &     &  15 & \Z.234375 &                    &     &    & 47 & \Z.734375\\
$2(\frac{1}{4}'')$ &    &     &     & \Z.25     & $6(\frac{3}{4}'')$ &     &    &    & \Z.75\\
                   &    &     &  17 & \Z.265625 &                    &     &    & 49 & \Z.765625\\
                   &    & \Z9 &     & \Z.28125  &                    &     & 25 &    & \Z.78125\\
                   &    &     &  19 & \Z.296875 &                    &     &    & 51 & \Z.796875\\
                   &  5 &     &     & \Z.3125   &                    &  13 &    &    & \Z.8125\\
                   &    &     &  21 & \Z.328125 &                    &     &    & 53 & \Z.828125\\
                   &    &  11 &     & \Z.34375  &                    &     & 27 &    & \Z.84375\\
                   &    &     &  23 & \Z.359375 &                    &     &    & 55 & \Z.859375\\
$3\phantom{(\frac{3}{8}'')}$ & & &  & \Z.375    & $7\phantom{(\frac{7}{8}'')}$ & & & & \Z.875\\
                   &    &     &  25 & \Z.390625 &                    &     &    & 57 & \Z.890625\\
                   &    &  13 &     & \Z.40625  &                    &     & 29 &    & \Z.90625\\
                   &    &     &  27 & \Z.421875 &                    &     &    & 59 & \Z.921875\\
                   &  7 &     &     & \Z.4375   &                    &  15 &    &    & \Z.9375\\
                   &    &     &  29 & \Z.453125 &                    &     &    & 61 & \Z.953125\\
                   &    &  15 &     & \Z.46875  &                    &     & 31 &    & \Z.96875\\
                   &    &     &  31 & \Z.484375 &                    &     &    & 63 & \Z.984375\\
$4(\frac{1}{2}'')$ &    &     &     & \Z.5      & $8(1'')$           &     &    &    &  1.0\rule[-2ex]{0pt}{0pt}\\
\hline\hline
\end{tabular}
\end{center}
%-----File: 097.png-----

\nbsubsection*{Metric System of Weights and Measures} %[**TN: original wording "Weights aud Measures"]
\index{Metric system of weights and measures}
\index{Measures, metric system of}

The metric system of weights and measures is
like the Arabic system of notation and the United
States money system in that it is based on the
decimal system; that is, its divisions and multiples
are all in the ratio of~$10$.

The metric unit of length is the \textit{meter} and all
other weights and measures of the metric system are
based on the meter. The meter is the one ten-millionth
part of the distance of the equator from
the pole. Its length is equivalent to $39.37$ inches.

The names of the different weights and measures
express their value, the Latin prefixes indicate the
fractional parts of the meter and the Greek prefixes
indicating multiples larger than the meter.
These prefixes are:

\nbtabletitle{Table XXI, Measure of Length}
\nbreflabel{table:XXI}
\index{Millimeter}
\index{Centimeter}
\index{Kilometer}
\index{Length, metric measure of}
\index{Metric measure of length}
\index{Measure of length, metric}

\nbtablesubtitle{(Common Table)}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l@{\;}c@{\;}l@{\;}c}
10 \textbf{millimeters} &=& 1 \textbf{centimeter} &=& \phantom{123,34}0.01 & meter\phantom{s}\\
10 \textbf{centimeters} &=& 1 decimeter           &=& \phantom{123,34}0.1  & `` \\
10 decimeters           &=& 1 \textbf{meter}      &=& \phantom{123,34}1    & `` \\
10 \textbf{meters }     &=& 1 decameter           &=& \phantom{123,4}10     & meters \\
10 decameters           &=& 1 hectometer          &=& \phantom{123,}100     & ``\\
10 hectometers          &=& 1 \textbf{kilometer}  &=& \phantom{12}1,000     & `` \\
10 \textbf{kilometers}  &=& 1 myriameter          &=& \phantom{1}10,000     & `` \\
10 myriameters          &=&                       & &           100,000     & ``
\end{tabular}
\end{center}

The denominations set in blackfaced type are
those in common use.
%-----File: 098.png-----

\nbtabletitle{Table XXII, Measure of Surface}
\index{Measure of surface, metric}
\index{Metric measure of surface}
\index{Surface, metric measure of}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l@{\;}c@{\;}c@{\;}c@{\;}l}
100 & square & \textbf{millimeters} &=& 1 & square & centimeter\\ %[**TN: only millimeters bold in original text]
100 &   ``   & centimeters          &=& 1 &  ``    & decimeter\\
100 &   ``   & decimeters           &=& 1 &  ``    & meter\\
100 &   ``   & meters               &=& 1 &  ``    & decameter\\
100 &   ``   & decameters           &=& 1 &  ``    & hectometer\\ %[**TN: original wording "hectometers"]
100 &   ``   & hectometers          &=& 1 &  ``    & kilometer
\end{tabular}
\end{center}

\nbtabletitle{Table XXIII, Measure of Land}
\index{Measure of land, metric}
\index{Metric measure of land}
\index{Land, metric measure of}

The unit of area is the \textit{are} and it is equal to one
square decameter.
\begin{center}
\begin{tabular}{r@{\;}c@{\;}c@{\;}r@{\;}l}
  1 & \textbf{are\phantom{s}} &=& 1000 & milliares, ma\\
  1 &                    ``   &=&  100 & centiares, ca\\
  1 &                    ``   &=&   10 & deciares, da\\
 10 &           \textbf{ares} &=&    1 & dekare, Da\\
100 &                    ``   &=&    1 & \textbf{hectare, Ha}
\end{tabular}
\end{center}

\nbtabletitle{Table XXIV, Measure of Volume}
\index{Cubic measure, of volume, metric}
\index{Volume, metric cubic measure of}
\index{Measure of volume, cubic, metric}
\index{Measure of volume, metric}
\index{Metric measure of volume}
\index{Volume, metric measure of}

\nbtablesubtitle{(Cubic Measure)}

The unit of volume is the \textit{cubic meter}.

\begin{center}
\begin{tabular}{@{}r@{\;}l@{\;}c@{\;}l}
1000 & cubic millimeters          &=& 1 \textbf{cubic centimeter}\\
1000 & \textbf{cubic centimeters} &=& 1 \textbf{cubic decimeter}\\ %[**TN: original wording "100 cubic centimeters"]
1000 & \textbf{cubic decimeter}   &=& 1 \textbf{cubic meter}
\end{tabular}
\end{center}

\nbtabletitle{Table XXV, Measure of Volume}
\index{Measure of wood by metric volume}
\index{Metric wood measure by volume}
\index{Volume, metric wood measure}
\index{Wood measure, metric volume}

\nbtablesubtitle{(Wood Measure)}

In measuring wood, excavations, and the like
a unit called the \textit{stere} is used and equals one cubic
meter.
%-----File: 099.png-----

\begin{center}
\begin{tabular}{r@{\;}c@{\;}l@{\;}c@{\;}r@{\;}l}
   1 & \textbf{stere},& s                     &=& 100& centisteres, cs\\
   1 &           ``   &                       &=&  10& decisteres, ds\\
  10 & \multicolumn{2}{@{}l}{\textbf{steres}} &=&   1& decastere, Ds\\
 100 &           ``   &                       &=&   1& hectostere, Hs\\
1000 &           ``   &                       &=&   1& kilostere, Ks
\end{tabular}
\end{center}

\nbtabletitle{Table XXVI, Measure of Volume}
\index{Liter}
\index{Liquids, metric measure by volume}
\index{Solids, metric measure by volume of}
\index{Measure of liquids by metric volume}
\index{Measure of solids by metric volume}
\index{Metric measure of liquids by volume}
\index{Metric measure of solids by volume}
\index{Volume, metric measure of liquid by}
\index{Volume, metric measure of solids by}

\nbtablesubtitle{(For Liquids and Solids)}

The \textit{liter} is the unit of capacity and is equal to
one cubic decimeter.

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
10 milliliters, ml              %[**TN: original wording "milliters"]
&=& 1 \textbf{centiliter}, cl\\ %[**TN: comma added before "cl"]
10 \textbf{centiliters} &=& 1 deciliters, dl\\
10 deciliters           &=& 1 \textbf{liter}, l\\
10 \textbf{liters}      &=& 1 decaliter, Dl\\
10 decaliters           &=& 1 \textbf{hectoliter}, Hl\\
10 hectoliters          &=& 1 kiloliter, Kl  %[**TN: "hectoliters" not bold in original text]
\end{tabular}
\end{center}

\nbtabletitle{Table XXVII, Measure of Weight}
\index{Weights, metric system of}
\index{Measure of weight, metric}
\index{Metric measure of weight}
\index{Weight, metric measure of}
\index{Gram}
\index{Kilogram}
\index{Quintal}
\index{Tonneau}

\begin{center}
\begin{tabular}{l@{\;}c@{\;}l}
10 milligrams, mg &=& 1 centigram, cg\\
10 centigrams     &=& 1 decigram, dg\\
10 decigrams      &=& 1 gram, g\\
10 grams          &=& 1 decagram, Dg\\
10 decagrams      &=& 1 hectogram, Hg\\
10 hectograms     &=& 1 kilogram, Kg\\
10 kilograms      &=& 1 myriagram, Mg\\
10 myriagrams     &=& 1 quintal, Q\\
10 quintals       &=& 1 tonneau, T
\end{tabular}
\end{center}
%-----File: 100.png-----

\nbtabletitle{A Key to the Metric System}
\index{Key to the twelve systems}
\index{Metric system, a key to the}

In the nickel five-cent piece of U.S.~coinage we
have a key to the metric system of weights and
measures. The diameter of this coin is $2$ centimeters
and its weight is $5$ grams. Five of these
coins placed in a row will, of course, have a length
equal to $1$ decimeter, and two of them will weigh
a decagram.

As the kiloliter is a cubic meter, a key to the
measure of length is also a key to the measure of
capacity. If therefore you will carry a five-cent
nickel for a pocket-piece you will always have a
key to the entire metric system of weights and
measures with you.

\nbsubsection*{Rules for Measuring Surfaces and Solids}
\index{Area}
\index{Measuring surfaces and solids, rule for}
\index{Rules for measuring surfaces and solids}
\index{Solids, rules for measuring}
\index{Surfaces, rules for measuring}
{\setlength\intextsep{0pt}
\setlength\columnsep{1.5em}
\begin{wrapfigure}{r}[-1.5em]{0pt}
\includegraphics[width=2.5in]{images/p93.pdf}
\end{wrapfigure}
\quad %dummy to let wrapfigure get started before actual paragraph

\nbsubparagraph{Parallelogram.}%
\index{Area of parallelogram, to find}%
\index{Parallelogram, to find area of}%
---To find the area of a parallelogram,
multiply its
length, or \textit{base} as it is
called, by its height, or
\textit{altitude} as it is called, or
expressed in the simple
form of an algebraic
equation.---
\[
A = b × h
\]
%-----File: 101.png-----

\begin{wrapfigure}[9]{l}[-1.5em]{0pt}
\includegraphics[width=1.5in]{images/p94a.pdf}
\end{wrapfigure}

\nbsubparagraph{Triangle.}%
\index{Area of triangle, to find}%
\index{Triangle, to find area of}%
---To find the area of a triangle when
the base and altitude are given,
multiply its base by its altitude
and divide by $2$, or
\[
A=\frac{bh}{2}
\]

\nbsubparagraph{Triangle.}---To find the area
when three sides are given take
half the sum of the three sides,
subtract each side separately from this sum; then
multiply half the sum of the three sides by the %[**TN: original wording "three sides of the"]
three remainders, and, finally, take the square root
of this product, thus:
\[
\begin{array}{l@{\qquad}l}
\text{\textit{First step}}&\dfrac{  a + b + c }{2}\\[0.9em]
\text{\textit{Second step}}&\dfrac{  a + b + c }{ 2} - a\\[0.9em]
\text{\textit{Third step}}&\dfrac{ a + b + c}{ 2 } - b\\[0.9em]
\text{\textit{Fourth step}}&\dfrac{a + b + c}{ 2}    - c
\end{array}
\]

\begin{wrapfigure}[7]{r}[-1.5em]{0pt}
\includegraphics[width=2in]{images/p94b.pdf}
\end{wrapfigure}
\quad %dummy to let wrapfigure get started before actual paragraph

\bigskip\nbsubparagraph{Square or Rhombus.}%
\index{Area of square, to find}%
\index{Square, to find area of}%
\index{Area of rhombus, to find}%
\index{Rhombus, to find area of}%
---The
area of a square or rhombus
are equal to the product of
the diagonals divided by 2, or
\[
A=\frac{ab}{2}
\]

\vspace{3\baselineskip} %to ensure figure 94b above starts high on the page
%-----File: 102.png-----

{%
\begin{wrapfigure}{r}[-1.5em]{0pt}
\includegraphics[width=2.9in]{images/p95a.pdf}
\end{wrapfigure}
\quad %dummy to let wrapfigure get started before actual paragraph

\smallskip
\nbsubparagraph{Trapezium.}%
\index{Area of trapezium, to find}%
\index{Trapezium, to find area of}%
---The area of a trapezium is equal
to half the product
of the sum
of the two parallel
sides by the
distance between
them, or
\[
A = \dfrac{(a + b) × c}{2}
\]
}

\nbsubparagraph{Circle.}%
\index{Area of circle, to find}%
\index{Circle, to find area of}%
---The area of a circle equals the square
of the diameter times .7854, or
\[
A = .7854 × d^2
\]

\bigskip
\begin{wrapfigure}{l}[-1.5em]{0pt}
\includegraphics[width=2.9in]{images/p95b.pdf}
\end{wrapfigure}
\nbsubparagraph{Ellipse.}%
\index{Area of ellipse, to find}%
\index{Ellipse, to find area of}%
---The area of an ellipse equals the
product of the
long and short
axes multiplied
by .7854, or
\[
A = a × b × .7854
\]
}% end \intextsep & columnsep values for wrapped figures

\bigskip
\nbsubsection*{Rules for Measuring Circles and Spheres}
\index{Algebraic formulas}
\index{Formulas, algebraic}
\index{Circles, rules for measuring}
\index{Spheres, rules for measuring}
\index{Measuring circles and spheres, rules for}
\index{Rules for measuring circles and spheres}

\nbsubparagraph{Circumference of a Circle.}%
\index{Circle, to find the circumference by algebra}%
\index{Circumference of a circle by algebra, to find the}%
---To find the circumference
of a circle multiply its diameter by
3.1416 or expressed by a simple algebraic formula---
\[
\pi D
\]
\begin{center}
Where $\pi$ (Greek letter pi) $= 3.1416$ and
$D =$ the diameter
\end{center}
%-----File: 103.png-----

\nbsubparagraph{Area of a Circle.}---To find the area multiply the
radius of the circle squared by 3.1416, or
\[
\pi R^2
\]

\nbsubparagraph{Surface of a Sphere.}%
\index{Surface of a sphere by algebra, to find}%
\index{Sphere by algebra, to find the surface of a}%
---To find the surface of a sphere
%[**TN: original wording "multiply the radius squared, or 3.1416, by the diameter squared by 12.5664"]
multiply the radius squared by 12.5664, or
the diameter squared by 3.1416
\begin{flalign*}
&\text{or} & & 4 \pi R^2 &\phantom{\text{or}} \\
&\text{or} & &   \pi D^2 &
\end{flalign*}

\nbsubparagraph{Volume of a Sphere.}%
\index{Volume of a sphere by algebra, to find the}%
\index{Sphere by algebra, to find the volume of a}%
---To find the volume of a
sphere multiply the radius of the circle cubed by
$\dfrac{12.5664}{3}$
\begin{flalign*}
&\text{or} & & \frac{3.1416}{6}× \text{ the diameter cubed}&\phantom{\text{or}}\\
&\text{or} & & \phantom{\frac{3.1416}{6}× \text{ th}} \tfrac{4}{3} \pi R^3\\
&\text{or} & & \phantom{\frac{3.1416}{6}× \text{ th}} \tfrac{1}{6} \pi D^3
\end{flalign*}

\nbsubsection*{Rules for Computing Simple Interest}
\index{Interest, rules for computing simple}
\index{Rules for computing simple interest}
\index{Simple interest, rules for computing}

The following are simple rules for finding the
interest on any principal for any number of days.
When the principal contains cents, point off four
places from the right of the result to express the
interest in dollars and cents. When the principal
contains dollars only point off two places.

{\setlength\parskip{4pt plus 2pt minus 1pt}
\setlength\parindent{0pt}
$2$ per cent, multiply the principal by the number
of days to run and divide by~$180$.

$2\frac{1}{2}$ per cent, multiply by number of days and
divide by 144.
%-----File: 104.png-----

$3$ per cent, multiply by number of days and
divide by $120$.

$3\tfrac{1}{2}$ per cent, multiply by number of days and
divide by $102.86$.

$4$ per cent, multiply by number of days and
divide by $90$.

$5$ per cent, multiply by number of days and
divide by $72$.

$6$ per cent, multiply by number of days and
divide by $60$.

$7$ per cent, multiply by number of days and
divide by $51.43$.

$8$ per cent, multiply by number of days and
divide by $45$.

$9$ per cent, multiply by number of days and
divide by $40$.

$10$ per cent, multiply by number of days and
divide by $36$.

$12$ per cent, multiply by number of days and
divide by $30$.

$15$ per cent, multiply by number of days and
divide by $24$.

$18$ per cent, multiply by number of days and
divide by $20$.

$20$ per cent, multiply by number of days and
divide by $18$.}
%-----File: 105.png-----

\nbchapter{IX}{Magic with Figures}
\index{Figures, magic with}
\index{Magic figures}

\begin{nbchaptercontentsone}
The Magic Squares\\
Second Sight with Dice\\
Telling a Person's Age\\
Thirteen Six Times in Twelve\\
How Figures Lie\\
1001 Different Deals with 13 Cards\\
Napier's Rods---A Mechanical Short Cut to Multiplication and Division\\ %[**TN: original wording "Multplication"]
Puzzle---Watchful Waiting\\
How to Become a Lightning Calculator in Ten Seconds
\end{nbchaptercontentsone}

\nbparagraph{Magic Squares.}%
\index{Magic squares}%
---The magic square is an arrangement
of numbers in the cells of a square in
such a fashion that however the rows are added,
whether vertically, horizontally, or diagonally, the
sum in every case will be the same.

Thus in the magic square here shown the numbers
\[
\setlength{\arraycolsep}{0.8em}
\renewcommand{\arraystretch}{2}
\begin{array}{|c|c|c|c|}
\hline
 1 & 16 & 11 &  6\\
\hline
13 &  4 &  7 & 10\\
\hline
 8 &  9 & 14 &  3\\
\hline
12 &  5 &  2 & 15\\
\hline
\end{array}
\]
from 1 to 16 are so disposed that whatever
%-----File: 106.png-----
row is added the sum is $34$. There are over $800$
different arrangements of possible number combinations
for forming magic squares and no two
will be alike.

Rules of various kinds have been worked out
for constructing magic squares but no single rule
has yet been found that is equally applicable for
all of them.

Magic circles\index{Magic circles} which consist of a number of concentric
circles, that is circles within circles, can be
formed by dividing their radii and numbering their
sections after the manner of the magic squares, and
when the numbers in each circle and row are added
the sum will be the same. \textit{Magic cubes}\index{Magic cubes} and \textit{magic
cylinders}\index{Magic cylinders} can also be constructed on the same
principle.

For a fuller account of Magic Squares see \textit{The
Monist} 1906--07, Vol.~XVI, page~56. Open Court
Pub.\ Co., Chicago.

Also the \textit{Occult Review}, Vol.~III, 1906. Published
by William Rider and Son Ltd., London. J.~M.~C. Hampson, 669 Sedgwick St., Chicago, Agent.

\nbsubsection*{Second Sight With Dice}\index{Second sight with dice}

\nbsubparagraph{The Effect.}---A pair o' dice are handed to some
one with the request that he throw them while
you turn your back or close your eyes or both so
that you cannot see the numbers.

Next ask him to take a number on either die
and to multiply it by $2$, then add $5$ and multiply
%-----File: 107.png-----
the sum obtained by $5$ and to this add the number
on the other die and have him tell you the number
he obtains from these operations.

From the number he gives you, you mentally
subtract $25$ and you will get two figures for the
remainder, and these will be the two numbers on
the dice.

\nbsubparagraph{Example:} Suppose the number given you is $59$;
then $59 - 25 = 34$, that is, the number on one die
is $3$ and the number on the other die is $4$. The
whole operation is as follows:
\[
3 × 2 = 6 + 5 = 11 × 5 = 55 + 4 = 59 - 25 = 34
\]

\nbsubsection*{Telling a Person's Age}
\index{Age, telling a person's}
\index{Telling a person's age}

\nbsubparagraph{The Effect.}---Ask a spectator to tell you in what
columns of figures his or her age appears in the
following table, when you will instantly tell his or
her age.

\nbsubparagraph{The Rule.}---By using the following table you
need only to know what columns his age appears
in and then add the top numbers of these columns
together, when the sum will be the age of the person
required.

\nbsubparagraph{Example:} Suppose a person says that his age
is given in columns A, C, and E, then---
\[
1 + 4 + 16 = 21
\]
hence the age value must be $21$.

The numbers of bank-notes, dates of coins, the
combinations of safes, the numbers of watches, the
%-----File: 108.png-----
%-----File: 109.png-----
time one gets up in the morning or any other
numbers thought of within the limits of the table
may be told by this method with equal facility.

\begin{center}
\vspace*{\stretch{1}}
\setlength{\tabcolsep}{2em}
\begin{tabular}{c|c|c|c|c|c}
\multicolumn{6}{c}{\normalsize TABLE\rule[-2ex]{0pt}{0pt}}\\
\hline
\hline
\rule{0pt}{3.5ex}A&B&C&D&E&F\rule[-2ex]{0pt}{0pt}\\
\hline
\rule{0pt}{3.5ex}\Z1&\Z2&\Z4&\Z8&16&32\\
\Z3&\Z3&\Z5&\Z9&17&33\\
\Z5&\Z6&\Z6&10&18&34\\
\Z7&\Z7&\Z7&11&19&35\\
\Z9&10&12&12&20&36\\
11&11&13&13&21&37\\
13&14&14&14&22&38\\
15&15&15&15&23&39\\
17&18&20&24&24&40\\
19&19&21&25&25&41\\
21&22&22&26&26&42\\
23&23&23&27&27&43\\
25&26&28&28&28&44\\
27&27&29&29&29&45\\
29&30&30&30&30&46\\
31&31&31&31&31&47\\
33&34&36&40&48&48\\
35&35&37&41&49&49\\
37&38&38&42&50&50\\
39&39&39&43&51&51\\
41&42&44&44&52&52\\
43&43&45&45&53&53\\
45&46&46&46&54&54\\
47&47&47&47&55&55\\
49&50&52&56&56&56\\
51&51&53&57&57&57\\
53&54&54&58&58&58\\
55&55&55&59&59&59\\
57&58&60&60&60&60\\
59&59&61&61&61&61\\
61&62&62&62&62&62\\
63&63&63&63&63&63\rule[-2ex]{0pt}{0pt}\\
\hline
\hline
\end{tabular}
\vspace*{\stretch{4}}
\end{center}

\nbsubsection*{Thirteen Six Times in Twelve}
\index{Thirteen six times in twelve}

To perform this curious operation in figures write
down the following numbers:
\[
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
\]
Now by adding the first and last numbers together
you will have---
\begin{align*}
1 +  12 &= 13\\
2 +  11 &= 13\\
3 +  10 &= 13\\
4 + \Z9 &= 13\\
5 + \Z8 &= 13\\
6 + \Z7 &= 13
\end{align*}

Thus with twelve numbers you get 13 six times.

\nbsubsection*{How Figures Lie}
\index{How figures lie}
\index{Figures, how they lie}

A young fellow had worked for a grocer for
seven years and got married, when it behooved him
to ask the boss for a raise; so the latter (who was
the original inventor of the wooden nutmeg) began
to do a little figuring.

``You see,'' he pointed out to his clerk, ``there
are $365$ days in a year and you work $8$ hours a day,
which makes $121\frac12$ days; then there are $52$ Sundays
and subtracting these leaves $69\frac12$ days. There are
%-----File: 110.png-----
$13$ legal, %[**TN: comma added]
$2$ race, and $1$ circus days, and taking these
away $54\frac12$ days are left.

``You have $1$ hour for lunch every day which,
summed up, makes $14$ days and this leaves $40\frac12$
days. Every Saturday afternoon you get off to
bury your grandmother or help a maiden aunt
hang pictures and this makes $25\frac12$ days and leaves
$14$ days. Then I have to give you a vacation of
two weeks every summer and this takes up the
other $14$ days. As the store is losing money I
can't give you the raise anyway. Good-night.''

\nbparagraph{1001 Different Deals with 13 Cards.}%
\index{Cards, 1001 different deals with, 13}%
---To ascertain
how many different deals can be made by
using $13$ cards out of an ordinary pack proceed as
follows:

First multiply the following terms: $52 × 51 ×
50 × 49 × 48 × 47 × 46 × 45 × 44 × 43 × 42 × 41
× 40$. You will note that there are $13$ terms in this
multiplication.

Next multiply the following $13$ terms: $13 × 12
× 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1$;
now divide the product of the first multiplication
by the product of the second multiplication and
the quotient will be the number of different deals
that can be made with $13$ out of a pack of $52$ cards.
If you know how to use \textit{logarithms} you can short-cut
the process very considerably.

\nbparagraph{Napier's Rods.}%
\index{Napier's rods}%
---A wonderful short-cut device
for multiplying and dividing was worked out three
centuries ago by Lord Napier. It has not been very
%-----File: 111.png-----
widely used since then in virtue of the fact that
\textit{logarithms}\index{Logarithms}, which he also invented, have taken its
place. It is however a very clever scheme and one
that you can make use of to good advantage
especially for the multiplication of large numbers.

\begin{center}
\sffamily
\renewcommand{\arraystretch}{1.6}
\begin{tabular}{|*{11}{@{}c@{}|}}
\multicolumn{1}{@{}c@{}}{\textbf{A}}\\
\hline
\nbnaptop{1}
&\nbnaptop{1}&\nbnaptop{2}&\nbnaptop{3}&\nbnaptop{4}&\nbnaptop{5}&\nbnaptop{6}&\nbnaptop{7}&\nbnaptop{8}&\nbnaptop{9}&\nbnaptop{0}\\
\hline
\nbnapside{2}
&\nbslash{\Z}{2}&\nbslash{\Z}{4}&\nbslash{\Z}{6}&\nbslash{\Z}{8}&\nbslash{1}{0}&\nbslash{1}{2}&\nbslash{1}{4}&\nbslash{1}{6}&\nbslash{1}{8}&\nbslash{\Z}{0}\\
\hline
\nbnapside{3} &\nbslash{\Z}{3}&\nbslash{\Z}{6}&\nbslash{\Z}{9}&\nbslash{1}{2}&\nbslash{1}{5}&\nbslash{1}{8}&\nbslash{2}{1}&\nbslash{2}{4}&\nbslash{2}{7}&\nbslash{\Z}{0}\\
\hline
\nbnapside{4} &\nbslash{\Z}{4}&\nbslash{\Z}{8}&\nbslash{1}{2}&\nbslash{1}{6}&\nbslash{2}{0}&\nbslash{2}{4}&\nbslash{2}{8}&\nbslash{3}{2}&\nbslash{3}{6}&\nbslash{\Z}{0}\\
\hline
\nbnapside{5} &\nbslash{\Z}{5}&\nbslash{1}{0}&\nbslash{1}{5}&\nbslash{2}{0}&\nbslash{2}{5}&\nbslash{3}{0}&\nbslash{3}{5}&\nbslash{4}{0}&\nbslash{4}{5}&\nbslash{\Z}{0}\\
\hline
\nbnapside{6} &\nbslash{\Z}{6}&\nbslash{1}{2}&\nbslash{1}{8}&\nbslash{2}{4} %[**TN: original wording "1/4"]
&\nbslash{3}{0}&\nbslash{3}{6}&\nbslash{4}{2}&\nbslash{4}{8}&\nbslash{5}{4}&\nbslash{\Z}{0}\\
\hline
\nbnapside{7} &\nbslash{\Z}{7}&\nbslash{1}{4}&\nbslash{2}{1}&\nbslash{2}{8}&\nbslash{3}{5}&\nbslash{4}{2}&\nbslash{4}{9}&\nbslash{5}{6}&\nbslash{6}{3}&\nbslash{\Z}{0}\\
\hline
\nbnapside{8} &\nbslash{\Z}{8}&\nbslash{1}{6}&\nbslash{2}{4}&\nbslash{3}{2}&\nbslash{4}{0}&\nbslash{4}{8}&\nbslash{5}{6}&\nbslash{6}{4}&\nbslash{7}{2}&\nbslash{\Z}{0}\\
\hline
\nbnapside{9} &\nbslash{\Z}{9}&\nbslash{1}{8}&\nbslash{2}{7}&\nbslash{3}{6}&\nbslash{4}{5}&\nbslash{5}{4}&\nbslash{6}{3}&\nbslash{7}{2}&\nbslash{8}{1}&\nbslash{\Z}{0}\\
\hline
\multicolumn{1}{@{}c@{}}{\textbf{A}}
\end{tabular}

\normalfont
\medskip
\small Fig.~1: Napier's Rods for Performing Multiplication Operations. %[**TN: "Fig. 1:" added (referred to in text)]
\end{center}

It consists of a set of ten rods having a square
cross-section and an index-rod AA as shown in Fig.~1.
This index-rod should be made stationary and it
can be fastened to a board. The other ten rods
should be loose so that they can be changed about.
%-----File: 112.png-----
A very serviceable set of rods can be made of strips
of heavy cardboard $\frac38$ inch wide and $3\frac34$ inches long.
These strips should then be divided off into squares
with the figures marked in them in good black ink.
Fig.~1 shows the set of rods or strips complete.

Now suppose that you want to multiply $3795$
by $9$; the first thing to do is to move the rods having
the numbers $3$, $7$, $9$, and $5$ at the top into the
position shown in Fig.~2, and up against the index-rod
AA\@. This done you are ready to perform the
operation, thus:
%-----File: 113.png-----

Follow the last figure of the multiplicand,
namely $5$, down until you are opposite the $9$ on
the index-rod which you will see in the lower half of
the square at $5$ and set this figure down in your
answer. Then add the $4$ and the $1$, that is, the
number in the upper part of the square to the
number in the lower part of the next square,
and since $4$ and $1$ are $5$ set this down in your
answer.

Adding, now, the number in the top of the next
square to the number in the lower part of the next
square, that is, $8 + 3$, you will have $11$, and set
down the unit figure ($1$) in the answer and carry
the tens figure ($1$) over to the next addition. Adding
as before you will have $6 + 7 = 13$ plus the
$1$ carried equals $14$.

Set down the $4$ in the answer and carry the
$1$, and since the figure in the upper part of the last
square is $2$ you add the $1$ carried to the $2$ which
makes $3$; set this down and you will get for your
answer $34155$.

\begin{center}
\sffamily
\renewcommand{\arraystretch}{1.6}
\begin{tabular}{|*{11}{@{}c@{}|}}
\multicolumn{1}{@{}c@{}}{\textbf{A}}\\
\hline
\nbnaptop{1}  &   \nbnaptop{3}&  \nbnaptop{7}&  \nbnaptop{9}&  \nbnaptop{5}\\
\hline
\nbnapside{2} &\nbslash{\Z}{6}&\nbslash{1}{4}&\nbslash{1}{8}&\nbslash{1}{0}\\
\hline
\nbnapside{3} &\nbslash{\Z}{9}&\nbslash{2}{1}&\nbslash{2}{7}&\nbslash{1}{5}\\
\hline
\nbnapside{4} & \nbslash{1}{2}&\nbslash{2}{8}&\nbslash{3}{6}&\nbslash{2}{0}\\
\hline
\nbnapside{5} & \nbslash{1}{5}&\nbslash{3}{5}&\nbslash{4}{5}&\nbslash{2}{5}\\
\hline
\nbnapside{6} & \nbslash{1}{8}&\nbslash{4}{2}&\nbslash{5}{4}&\nbslash{3}{0}\\
\hline
\nbnapside{7} & \nbslash{2}{1}&\nbslash{4}{9}&\nbslash{6}{3}&\nbslash{3}{5}\\
\hline
\nbnapside{8} & \nbslash{2}{4}&\nbslash{5}{6}&\nbslash{7}{2}&\nbslash{4}{0}\\
\hline
\nbnapside{9} & \nbslash{2}{7}&\nbslash{6}{3}&\nbslash{8}{1}&\nbslash{4}{5}\\
\hline
\multicolumn{1}{@{}c@{}}{\textbf{A}}
\end{tabular}

\normalfont
\medskip
\small Fig.~2: Rods Set to Multiply $3795$ by $9$. %[**TN: "Fig. 2:" added (referred to in text)]
\end{center}

Sets of numbers can be multiplied just as easily
as the simple example given above, for all you
have to do is to slide the rods or strips on which the
figures of the multiplicand are marked next to the
index-rod.

Napier also devised a division table and you
can make one by means of a table of the product
of the divisor multiplied by each of the numbers
from $1$ to $9$.
%-----File: 114.png-----

\nbparagraph{Puzzle---Watchful Waiting.}%
\index{Puzzle---watchful waiting}%
---\nbsubparagraph{The Question.}---At
what points on the dial of a watch or clock will
the hour and minute hands be in conjunction
during $12$ succeeding hours if they are started at
exactly 12~o'clock?

%[**TN: Answers are accurate to the nearest half-minute, but not exact]
\nbsubparagraph{The Answer.}---At the following times during
one complete revolution of the hour hand:
\[
\begin{tabular}{r@{}l@{\;}c@{\;}c@{\;}l}
At&$\Z5\frac12$ &minutes&after&\Z1\\
  &$ 11$        &  ``   & ``  &\Z2\\
  &$ 16\frac12$ &  ``   & ``  &\Z3\\
  &$ 22$        &  ``   & ``  &\Z4\\
  &$ 27\frac12$ &  ``   & ``  &\Z5\\
  &$ 33$        &  ``   & ``  &\Z6\\
  &$ 38$        &  ``   & ``  &\Z7\\
  &$ 43\frac12$ &  ``   & ``  &\Z8\\
  &$ 49$        &  ``   & ``  &\Z9\\
  &$ 54\frac12$ &  ``   & ``  & 10\\
  &$ 60$        &  ``   & ``  & 11
\end{tabular}
\]

\nbparagraph{How to Become a Lightning Calculator in Ten
Seconds.}%
\index{Lightning calculator, how to become one in 10 seconds}%
---This is a pretty trick with figures that
has been used by many a lightning calculator (?)
as a preliminary to the further exploit of distributing
Yellowstone Kit's two dollar liver-pad for the
small sum of one dollar.

\nbsubparagraph{The Effect.}---You ask a spectator to write
down a row of figures and then you write down a
row; have him write another row and you write
one under it; he writes a third row and you
your third row, and so on without regard to the
%-----File: 115.png-----
number of figures in the row or to the length of the
column. Then you write down the sum as fast
as you can chalk the figures on the board.

\smallskip
\nbsubparagraph{An Example:}
\begin{center}
\begin{tabular}{l@{\;}c@{\quad}r@{}}
First  &row&65421\\
Second &`` &34579\\
Third  &`` &52578\\
Fourth &`` &47422\\
Fifth  &`` &97351\\
Sixth  &`` & 2649\\
Seventh&`` &23451\\
Eighth &`` &12345\\\cline{3-3}
&&335796
\end{tabular}
\end{center}

\nbsubparagraph{The Cause.}---You will observe by referring to
the above example that the corresponding figures
of the first and second rows make $9$ when added
together, except the right hand ending figures
which when added makes $10$, thus:
\begin{center}
\begin{tabular}{l l l l r}
6&5&4&2&1\\
3&4&5&7&9\\
\hline
9,&9,&9,&9,&10\\
\end{tabular}
\end{center}

And this also holds good for the third and fourth
rows and for the fifth and sixth rows. The seventh
and eighth rows however can be written either by
the spectator or by yourself, and \textit{any figures} may be
used as long as they are \textit{less} than $9$.

To add the column start at the left and set
down $3$, since there are three sets of rows which
%-----File: 116.png-----
you have artfully \textit{doped out}. Then simply add the
last two rows, setting down the figures in their
proper order, thus:
\begin{center}
\begin{tabular}{l l l l l l}
\Z&2&3&4&5&1\\
\Z&1&2&3&4&5\\
\hline
3&3&5&7&9&6\\
\end{tabular}
\end{center}
and this is the answer.

Of course there is no real adding and the last
two rows can be written anywhere in the example
as long as the rows forming the couplets add to $9$
with the two last figures adding to $10$.
%-----File: 117.png-----
%-----File: 118.png-----
\cleardoublepage
\thispagestyle{empty}
\begin{center}
\nbvspace
\Large \textbf{INDEX}\\[5em]
\nbvspace
\end{center}
\cleardoublepage
%-----File: 119.png-----
%-----File: 120.png-----

% Bookmark the index for ease of navigation, even though it is not in ToC
\pdfbookmark[0]{Index}{Index}

% no header rule from beginning of index to end of book
\renewcommand\headrulewidth{0pt}

\printindex
%-----File: 121.png-----
%-----File: 122.png-----
%-----File: 123.png-----
%-----File: 124.png-----
%-----File: 125.png-----
%-----File: 126.png-----
%-----File: 127.png-----
%-----File: 128.png-----

\clearpage
\pagestyle{empty}
\nbreflabel{section:notef}
\textit{Explanation of \hyperref[section:rulef]{Rule F}}, page~\pageref{section:rulef}:

\medskip
This rule applies only when it is possible to subtract
\textit{arithmetically} and \textsc{not algebraically}, the sum of the
even place figures from the sum of the odd place figures.

\clearpage

\pagestyle{empty}
\vspace*{\fill}
\pdfbookmark[0]{Transcriber's Notes}{Transcriber's Notes}

{%
  \setlength{\parindent}{0pt}
	\setlength{\parskip}{6pt plus 2pt minus 1pt}
  \subsection*{{\normalsize\centering\itshape Transcriber's Notes}}
  \label{section:tn}
  \small

  Typographical and arithmetic errors and inconsistencies have been corrected,
  and are documented in the \LaTeX\ source as follows:
  {\ttfamily\footnotesize\%[**TN: text of note]}

  \hyperref[table:I]{Table I}, The Roman Numerals, has had several errors corrected,
  while attempting to preserve the author's intentions.

  \hyperref[table:III]{Table III}, Other Measures of Length, describes the \textit{knot}
  as a measure of length, when it is actually a measure of speed (one nautical mile per hour).

  The final column in \hyperref[table:XXI]{Table XXI}, Measure of Length, has been corrected
  by multiplying all of its entries by ten.

  \hyperref[section:rulef]{Rule F}, given for divisibility by seven, is not accurate,
  nor does the \hyperref[section:notef]{explanatory note} give a rule which is
  true in general.

  \TransNoteText
}

\cleardoublepage

\backmatter

\pdfbookmark[0]{Project Gutenberg License}{Project Gutenberg License}

\pagestyle{fancy}
\markboth{LICENSING}{LICENSING}
\pagenumbering{Alph}

\begin{verbatim}
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