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·Figuring The Joy of Numbers There is a sange and nichness to numbers they can come alne crase to be symbols written on a black boud and lead the reader into a world of intellectual dienture where alculations are thrilling In Figuring Shalunula Deu dramatizes the endless fascination ol numbers ind their ability to amaze and enitytain She oſſers easy to learn short cuts-hou to add long columns in your he id multiply drude and find square roots quickly almost magically Fraccions decimals and compound interese become clear and easy to deal with The author takes delighun working out huge problems mentally-sometimes even faster than lectromc computers In Figuring she shares her secrets sith you Shakuntala Dou shous you a loolproof method to guess your friends butihdays and how to determine the day of the week for any dare since 1582 and sublimely ridiculous she proves that 1= 2 Puzzle lovers wall enjoy the stories of the wise temple prest who divided 17 elephanis in half without bloodshed the man who will be 3 years older next year than he was the day before yesterday and the ſeat of describing each number from { through 18 with exactly 4 fours in cach equation Shakuntala Devi has delighted world audiences with her lightning fast computations and her infectious enthusiasm and love for numbers No one can read this book without falling under the spell of Figuring and or Ms Shakuntala Devi

·By the same author
in
Orient Paperbacks
Puzzles to Puzzle You
Vore Puzzles to Puzzle You
The Book of Numbers
Astrology for You
Perfect Murder

·FIGURING
The Joy of Numbers
Shakuntala Devi
ORIENT PAPERBACKS
A Division of Vision Books Private Limited

·Ise Published in Orient Paperbacks 1996
Reprinted 1987
Figuring The Joy of Numbers
Shakuntala Devi
Cover Designed by Vision Studio
Published by
Orient Paperbacks
(1 Division of Vision Books Pvt Lid )
Madarsa Road hashmere Gate Delhi 110006
Printed in India
Vladan Hall Tone Co Delhi 110006
inted at
Printing Press Delhi 110006

·CONTENT'S Introduction 1 Some Terms Defined 11 2 The Digits 15 31 3 Muluplication 4 Addition 48 5 Division 52 6 Subtraction 60 7 GCM and LCM 62 68 8 Squares and Square Roots 9 Cubes and Cube Roots 10 Percentages Discounts and Interest 77 87 11 Decimals 94 12 Fractions 99 104 13 The Calendar 14 Some Special Numbers 212 15 Tricks and Puzzles 131 Conclusion 157

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·INTRODUCTION" At three I fell in love with numbers It was sheer ecstasy for me to do sums and get the right answers Numbers were toys with which I could play In them I found emotional security, two plus two always made, and would always make, four - no matter how the world changed My interest grew with age I took immense delight in working out huge problems mentally - sometimes even fas er than electronic calculating machines and computers I travelled round the world giving demonstrations of my talents in every country I performed for students, professos, teachers, bankers, accountants, and even lay men who knew very little or nothing at all, about mathematics These performances were a great success and everywhere I was offered encouragement and appreciation All along I had cherished a desire to show those who think mathematics boring and dull just how beautiful it can be This book is what took shape It is written in the knowledge that there is a range and richness to numbers they can come alive, cease to be symbols written on a blackboard, and lead the reader into a world of intellectual adventure where calculations are thrilling not tedious To understand the methods I describe you need nothing more than a basic knon ledge of arithmetic There are no formulae technical terms algebra, geometry, or logarithms here - if you 9 ·10 Introduction can carry out the basic functions of adding, subtracting, mulu plying, and dividing you can follow all of it At first glance some of the steps may seem complicated but if you take them slowly you should be able to acquire the skill to solve a wide variety of problems without mistakes - and to do this mentally, or with a minimum use of pencil and paper But above all I would wish this book to give people an appreciation of the enchantment of numbers ·SOMC 7 CRAS DININED In this book *c are not going to be concerned with loganthms or the Calculus not even with algebra, cxcept in the most clementary tras So those who found that these aspects of mathematics gaic them a life-long terror of the whole subject an relax We are going to be dealing with ordinary numbers which should hold no terrors for anyone We will be encountering some mathematical terms and conventions but most of them arc in everyday use. I shall be using them because thes are the simplest and clearest means of stating mathemaucal problems or explananons But they do all have precise meanings and even if you are confident that you understand them, it would be as well to read this short chapter so that he can both be sure we are talling about the same things A Digit is an indırdual numeral there are ten of them 1,2,3 4 5,6 7, 8 9 and 0 A Fraction in every day language means a part of a whole I mathematics it normally has a slightly more precise meanin a part of a s hole expressed by putting one figure oser anoth dividing them by a line as in or The figure on top is 11 ·12 Figuring the numerator (it tells you how many parts you sc got) and the onc hcles is called the dennınır stur (it tells you what proportion of a whole each part is) They can also be written as 1/2 or 1/4 where the obliquc sirokc desides numcrator and denominator of expressing the same Decimals arc simply a different 153) thins, Instead of the denominator being variable as in fractions, in decimals it is always ten or rather tons because the system can be extended infinitely to hundredths, thousandths tenthousandths, ctc of a whole Thus 1 is the same as 1/10 and 01 is the same as 1/100 and 11 is the same as 11/100 Percentage is yet another way of cxpressing the same thing This timr the denominator is always 100 - so 4% is the same as 04 or 4/100 Percentages can be combined with decimals or fractions if it is necessary to express parts smaller than onehundredth Thus onc-five-hundredth can be cxpressed either as 2% or as % Multiplication is of course to multiply one number by another The number that is multiplicd is called the multiplicant, the number by which it is to be multiplied is called the multspher and the result is called the product of the two numbers Thus 2 (multiplier) 4 (multiplicand) = 8 (product) A multiple of a number is simply a number which is made up by multiplying the original number a given number of times Thus 2 4 6, 8 etc are all multiples of 2 Division is the reverse of multiplication, the division of one number by another The number to be divided is called the drordend and the number by which it is to be divided is called the dnisor If the divisor does not go into the dividend an exact umber of times then the figure left over is the remainder For ·example 9 (dividend) - 2 (divisor) with leftovers remainder If there is no remainder then the resuldanal divisor are factors of the dividend For instance, 2 and are factors of 6 The factors of a number can also be broken down into their smallest parts, thus the factors of 12 are 2, 2 and 3, or as it is usually put, 2 x 2 x 3 Addition and subtraction are so basic that all we need note is that in addition the result is known as the sum of the numbers that have been added up, and in subtraction the result is known as the remainder, 1e, what is left over after one number has been taken from another Roots and powers If a number is multiplied by itself, the prom duct is the square of that number - the square of 2 is 4 and twosquared is written 2 If the square is again multiplied by the original number, le 22 x 2, then the result is the cube of the number, which is written 23 The process can of course be continued ad infinitum - and is called raising a number to its fourth or fifth or 6785th power The root of a number is the original number which was squared or cubed or raised to the power of somethung in order to produce that number For example the square root of 9 is 3 because 3 x 3=9 The cube root of 27 is also 3 because 3 x 9 = 27, and the fourth root of 81 is also 3 because 3 x 27=81 The way this is written is 79 in the case of a square root and 8/27 or 781 in the case of cube, fourth or further roots The sixth root of 64 is 2, so 2 can be written mathematically as 8/64, just as 64 can be written 2° We will be encountering a few more technical terms as we go along, but these are the basic ones which you will need right through the book It s important that, however elementary they seem, you fully grasp what they mean As I said above they do ·14 Figuring have precise meanings - for example the difference between the product of 4 and 4 (4 x 4 = 16) and the sum of 4 and 4 (4 + 4 = 8) is obviously scry important, so it is worth just rechecking to make sure that they are all absolutely clear in your head before going any further

·2
THE DIGITEEN
The ten digits, 1, 2, 3, 4, 5, 6, 7, 8,9 2000 ze the 'alpha hag ca
arithmctic Arranged according to the corrections of me.be
matics and in conjunction with sobi's moon, sub.s.-XL
etc the digits make up'sentences phasesuauo-s 3-3.5
solutions Just as the letters of the fire their riorem
liarities and rules ( before e consec), see
each have a particular chansetic Snake CE
oddities can be put to good es, este recomanda
amusing for their own sake.

·16 Figuring sum of the digits comes to a ino-digit number (fnt cxamplc 456, 4+ 5+ 6 = 15) then normally you repeat the process (1 + 5 = 6) The secret steps in the multiplication tables are made bu applying this technique to the products glen by the strai, he steps I or instance in the 4 times tablc 3 y 4 = 12 (strairht step) 1 +2 = 3 (sccrct step) or in the 9 times table 11 x 9 = 99,9 + 9 = 181 +8=9 So with that techniquc in mind, let us lousk at the dists onc at a time NUMBERT The first thing about the number 1 is of course that whatever figure you multiply it by, or divide by it remains unchanged The first nine places in the I times table thercforc have no sccret steps or at least they are the same as the normal ones, but after that it becomes more interesting 10 x1 = 10 11 XI = 11 12 x1 = 12 13 x1 = 13 14 x1 = 14 15 XI = 15 16 x1 = 16 1+0=1 1+1=2 1+2 = 3 1 +3=4 1+4= 5 1+5=6 I+6=7 and so on No matter how far you go you will find that the secret steps always give you the digits from 1 10 9 repeating themselves in scqucnct For example ·The Digits 17 Secret steps Straight steps 154 x1= 154 155 x 1= 155 156 x1= 156 157 x1= 157 158 x 1= 158 159 XI = 159 160 x1= 160 161 x 1 = 161 1 +5+4= 10,1 +0= 1 1 +5+5=11 1+1=2 1 +5 +6= 12,1 +2=3 1+5+7= 13,1 + 3 = 4 1 +5+8= 14 1 + 4 = 5 1 +5 +9= 15 1 +5=6 1+6+0 1+6+1 8 = 7 In this case it is fairly obvious why the secret steps work but none the less it ... a curiosity that may not have occurred to you before Another curiosity about the number 1 is its talent for creating palindromes (numbers that read the same backwards as forwards) 1x1 11 x 11 111 x 111 1111 x 1111 11111 x 11111 111111 x 111111 1111111 x 1111111 11111111 x 11111111 111111111 x 111111111 1 121 12321 1234321 123454321 12345654321 1234567654321 123436787654321 12345678987654321 At that point it stops but the same thin, works briefly with the number 11 11 11 121 Il xil x 11 = 1331 11 x 11 x 11 x 11 =14641 ·18 Figuring Finally the fact that I 15, so to speak, immune to multiplication means that whatever power you raise it to it remains unchanged 1641746 = 1 By the same token, V1 = 1 NUMBER 2 The number 2 has one very obvious characteristic to multply any number by 2 is the same as adding it to itself As for the multiplication table and its secret steps 2*1 = 2 2 x2= 4 2 x 3 - 6 2 x 4 - 8 2 4 6 8 2 X 510 2*6 =12 2 * 7 = 14 2 x 8 = 16 2 x 918 2 x 10 = 20 2 x 11 = 22 2 x 12 =- 24 2 y 13 =26 2 x 14 = 28 2 x 15 = 30 2 x 16 = 32 2 x 17 = 34 2 x 18 = 36 1+0=1 1 +2 = 3 1+4= 5 1+6=7 1 + 8 - 9 2 +02 2+2=4 2 + 4 = 6 2 +6=8 2 +8= 10) 3 +0=3 3 + 2 = 5 3+4=7 3+6=9 And so on ·The Digits 19 As you can check for yourself, the secret steps go on working ad infinitum, always giving you the same sequence of the four even digits followed by the five odd ones There is an amusing little party trick that can be played with the number 2 The problem is to express all ten digits, in each case using the number 2 five times and no other number You are allowed to use the symbols for addition, subtraction, multiplication and division and the conventional method of writing fractions Here is how it is done 2+2 -2 -2/2 = 1 2 + 2 +2 -2 -2=2 2+2–2 + 2/2 = 3 2 x 2 x 2 -2 -2=4 2+2+2 - 2/2 = 5 2 + 2 + 2+2-2=6 22 -2 -2 -2=7 2x2x2 +2 -2=8 2 x 2 x 2 + 2/2 = 9 2 - 2/2 - 2/2 = 0 Finally, here is another oddity assnciated with 2 123456789 + 123456789 + 987654321 + 987654321 +2 2222222222 ·20 Figurmg NUMBER 3 The number three has two distinctions First of all it is the first triangle number - that is a number the units of which can be laid out to form a triangle, like this o' Triangle numbers have importance and peculiarities of their own which we shall encounter later on Three is also a prime number, a number that cannot be evenly divided except by itself and by 1 1 and 2 are of course prime numbers the next after 3 is 5, then 7, 11, 13, 17, 19 and 23, after that they gradually become increasingly rare There are a couple of strange things about the first few prime numbers, for example 153 = 18 + 5% + 38 And 3 and 5 can also both be expressed as the difference between two squares 3 = 2 - 12 5 = 32 - 2* The secret steps in the 3-times table are very simple Secret steps Straight steps 3 x1 = 3 3 X 2= 6 x3 = 9 3 6 9 3 x 4 = 12 3 x 5= 15 3 x6 =18 1 +2=3 1+5=6 1+8= 9 ·The Digits 21 3 x 7 21 3 x 8 = 24 3 x 9 = 27 2+1=3 2+4= 6 2+7=9 3 x 10 = 30 3 X 11 33 3 x 12 = 36 3+0=3 3 + 3 = 6 3 +6=9 Again the pattern of the secret steps recurs whatever stage you carry the table up to - try it and check for yourself NUMBER 4 With number 4 the secret steps in the multiplication tables become a little more intricate Secret steps 4 3 2 2+8= 10 1 Straight steps 4X14 4 x 2 = 8 4 x3 =12 4 X 4 = 16 4 X 5 = 20 4 y 6= 24 4 X 7=28 4 x 8 = 32 4 x 9 = 36 4 x 10 = 40 4 X 11 = 44 4 x 12 = 48 4 x 13 = 52 4 x 14 = 56 4 x 15 = 60 4 x 16 = 64 8 1+2= 1 +6=7 2+0= 2 + 4 = 6 1+0= 3+2=5 3+6 4+0=4 4 +4= 1+2=3 5+2= 1+1=2 6 +0= 1+0=1 9 8 4 + 8 = 12 7 5+6= 11 6 6 + 4 = 10 ·22 Figurmg At first the sums of the digits look lke a jumble of figures, but choose at random any sequence of numbers and multiply them by 4 and you will see the pattern emerge of two interlinked columns of digits in descending order for cample 2160 X4=8640 8+6+4+0=18 1+8= 9 2161 x 4=8644 8+6+4+4=22 2+2=4 2162 x 4=8648 8+6+4+8=26 2+6= 8 2163 x 4=8652 8+6+5+2=21 2+1=3 2164 x 4 = 8656 8+6+5+6=25 2+5= 7 2165 x 4=8660 8+6+6+0=20 2+0=2 2166 x 4=8664 8+6+6+4=24 2+4= 6 2167 x 4 =8668 8+6+6+-8=28 2+8=10 1+0=1 2168 x 4=8672 8+6+7+2=23 2+3= 5 NUMBER 5 Perhaps the most important thing about 5 is that it is half of 10, as we will see this fact is a key to many shortcuts in calculation In the meantime the secret steps in the 5-times table are very similar to those in the 4-times, the sequences simply go upwards instead of downwards Secret steps 5 6 Straight steps 5 x1= 5 5 X 2= 10 5 x 3 = 15 5 X 4 = 20 5 x 5 = 25 5 X 6 = 30 5x7=35 1+0=1 1+5= 2+0=2 2+5= 3 +0=3 3+5= 7 8 ·RI Brand 7 The Digsis 23 9 5 +5= 10 1 6+5=11 2 5 x 8 = 40 5 x 9 = 45 S X 1050 5 x 1l = 55 5 x 12 = 60 5 x 13 = 65 5 x 14 = 70 5 x 15 = 75 5 x 16 = 80 5 x 17 85 5 x 18 = 90 5 x 19 = 95 4 +5= 5+0=5 1+0= 6 +0a6 1-41- 7+0=7 1+2= 8 +0=8 I th3 = 9 +0=9 7+5 = 12 بیا 8 + 5 = 13 4 9+5= 14 And so on 9992 prato. NUMBER 6 28.4 88 This is the second triangle number, andrahe first performand a perfect number is one which is equal to the sum of ali divisors Thus, 1+2+3=6 The secret steps in the 6-times table are very similar to thosc in the 3-times, only the order is slightly different Serret steps Straight steps 6x1= 6 6 x 2 = 12 6x3 = 18 6 X 4 = 24 6 X 5 = 30 6x6 = 36 6 x 742 6 X 848 6 1+2=3 1+8=9 2+4=6 3+0=3 3+6=9 4+26 1+2-3 4+8 = 12 ·24 Figuring 6x9 = 54 6 x 10 = 60 6 x 11 = 66 6 x 12 = 72 6 + 6 = 12 5+4= 9 6 +0= 6 1+2=3 7+2=9 And so on NUMBER 7 This is the next prime number after 5 The secret steps in the 7-times table almost duplicate those in the 2-times, cxcept that they go up instead of down at ach step Straight sleps Secret steps 2 + 8 = 10 4+9=13 5+6= 11 7x= 7 7 x 2 = 14 7 x3 = 21 7*4 = 28 7x5= 35 7 x6 = 42 7x7 = 49 7x8= 56 7x9 = 63 7 x 10 = 70 7 x 113 77 7 x 12 = 84 7 x 13 = 91 7 x 14 = 98 7 x 15 = 105 7 x 16 = 112 7 x 17 = 119 7 x 18 = 126 7 1 *4 = 5 2+1=3 1+0=1 3 + 5 = 8 4+2= 6 1+3=4 1+1=2 6 + 3 = 9 7+0= 7 1+ 4 = 5 1+2= 3 1+0=1 1+7=8 1+0+5=6 1+1+2=4 1+1=2 1 +2 +6=9 7+7= 14 8 + 4 = 12 9+110 9 +8=17 1+1 +9 = 11 ·7 x 19 = 133 7 x 20 = 140 The Digits 25 1 +3+3=7 1 +4+0=5 And so on. There is a curious relationship between 7 and the number 142857 Watch 14 . 7x2 - 7x2 7 x2 = 7x4 7 x 28=7X8 7 x 24 =7 16 - 7 x 28 57 x 32 = 7 x 26 ==7 x 64 = 7 x 27=7 x 128 = 7 x 28=7 x 256 = 7 x 2° = 7x 512 = 28 56 112 224 448 896 1792 3584 142857142857142(784) However far you take the calculation, the sequence 142857 will repeat itself, though the final digits on the right-hand side which I have bracketed will be 'wrong' because they would be affected by the next stage in the addition if you took the calculation on further This number 142857 has itself some strange properties, multiply it by any number between 1 and 6 and see what happens 142857 X 1 = 142857 142857 x 2 = 285714 142857 X 3 = 428571 142857 X 4 = 571428 142857 x 5=714285 142857 x 6=857142 ·26 Figurmg The same digits recur in each answer, and if the products are each written in the form of a circle, you will sce that the order of the digits remains the same If you then go on to multiply the same number by 7, the answer iS 999999 e mill come back to some further characteristics of this number in Chapter 14 NUMBER 8 This time the secret steps in the muluplıczulon table are the reverse of those in the l-times table Secret steps Straight steps 8x1 - 8 8 X 2= 16 8 x 3 = 24 8 X 4 = 32 8 x 5 = 40 8 X 6 = 48 8 x 7 = 56 8 x8= 64 8 X 9 72 8 x 10 = 80 8 x 11 = 88 8 x 12 = 96 8 x 13 = 104 8 X 14 = 112 8 x 15 = 120 8 x 16 = 128 8 x 17 = 136 8 1+6=7 2 +4 = 6 3+2 = 5 4+0=4 4 + 8 = 12 1 +2=3 5+6= 11 1+1=2 6 + 4 = 10 1+0=1 7+2=9 8+0= 8 8+8= 16 1 +6=7 9 +6 = 15 1+5=6 1+0+4-5 1+1+2=4 1+2+0=3 1+2+8= 11 1+1=2 1+3+6= 10 1+0=1 So it continues

·The Digsis 27
If this is unexpected, then look at some other peculiarities of
the number 8
888
88
8
8
8
and
1000
88 9 x 9 +7
888 =98 x 9 +6
8888 = 987 x 9 + 5
88888 9876 x 9 + 4
888888 = 98765 x 9 +3
8888888 987654 x 9 +2
88888888 = 9876543 x 9 +1
and, lastly
12345679 x 8 98765432
NUMBER 9
With 9 we come to the most intriguing of the digits, indeed.
of all numbers Furst look at the steps in the mulap
table
$

·Secret steps 28 Figurmg Straight steps 9x1 9 9 9 x2 = 18 1+8=9 9 X 3 = 27 2+7==9 9 X4 = 36 3+6 = 9 9X5 = 45 4+5=9 9 x6 = 54 5+409 9 x 7 = 63 6+3=9 9 x 8 = 72 7+2=9 9 X 9 81 8+1=9 9 x 10 = 90 9 +0=9 9 x 11 = 99 9 +9 = 18 1 +8=9 9 x 12 = 108 1+0+8=9 It is an absolute rule that whatever number you multiply by 9, the sum of the digits in the product will always be 9 Morcover, not only are there no steps in the hidden part of the S-times table but, for its first ten places, it has another feature 1 x9=09 2 x 9 = 18 3 x 9 = 27 4 x 9 = 36 5 x 9 = 45 90 9 x 10 81 =9x9 72 = 9 x 8 63 = 9x7 54 = 9x6 The product in the second half of the table is the reverse of that in the first half Now, take any number Say, 87594 Reverse the order of the digits which gives you 49578 Subtract the lesser from the greater 87594 -49578 38016

·The Digits 29
Now add up the sum of the digits in the remainder 3 + 8 +
0+1+6 = 18,1 +8=9 The answer will always be 9
Again, take any number Say, 64783 Calculate the sum of its
digits 6+4+7+8+3 = 28 (you can stop there or go on
as usual to calculate 2 + 8 = 10, and again, only if you wish,
1+0=1)
Now take the sum of the digits away from the original number,
and add up the sum of the digits of the remainder Wherever
you choose to stop, and whatever the number you originally
select, the answer will be 9
64783
-28
64755
6+4+7+5
vs Here
TR299
64783
--10
64773
6+4+7+7+3 = 27 ng
64783
64782
6+4+7+8+2=27
2+7=9
Take the nine digits in order and remove the 8 12345679
Then multiply by 9
12345679 X 9= 111111111
Now try multiplying by the multiples of 9

·30 Figumg 12345679 x 18 = 222222222 12345679 x 27 = 333333333 12345679 X 36 = 44444444 12345679 x 45 = 55555555. 12345679 X 54 = 66666666 12345679 x 63 77777777 12345679 X 72 888888888 12345679 x 81 = 999999999 There is a little mathematical riddle that can be played using 9 What is the largest number that can be written using three digres? The answer is 95", or 9 raised to the ninth power of 9 The ninth power of 9 1$ 387420489 No one knows the precise number that is represented by 9387420489, but it begins 428124773 and ends 89 The complete number will contain 369 million digits, would occupy over five hundred miles of paper and take years to read ZERO I have a particular affection for zero because it was some of my countrymen who first gave it the status of a number Though the symbol for a void or nothingness is thought to have been invented by the Babylonians, it was Hindu mathematicians who first conceived of 0 as a number, the next in the progression 4-3-2-1 Now, of course, the zero is a central part of our mathematics, the key to our decimal system of counting And it significs something very different from simply nothing - just think of the enormous difference between 001, 01, 1, 1, 10, and 100 to remind yourself of the importance of the presence and position of a 0 in a number The other power of O is its ability to destroy another number - zero times anything is zero ·3 MULTIPLICATION Now that we have defined the principal terms we are going to use and have had a look at the digits individually, we can move on to look at some actual calculations and the ways in which they can be simplified and speeded up I do want to stress at this point that I am not suggesting any magical method by which you can do everything in your head Do not be ashamed of using paper and pencil, especially when you are learning and practising the different methods I am going to explain At first you may find you need to scribble down the figures at each intermediate stage in the calculation, with practice you will probably find that it becomes less necessary But the point is not to be able to do without paper, it is to grasp the methods and understand how they work Everyone can do simple multiplication in thex head All mothers are familiar with questions like 4 sausages each for 5 children, or the weekly cost of 2 pints of milk a day at 9 pennies a pint Problems usually arise when both the figures get into two or three digits - and this is, in many cases, because of the rigid methods we were taught at school If for example you multiply 456 by 76 by the method usually taught in schools, you end up with a calculation that looks like this 31

·34 Figures
87 60 x 100 = 8760
8760 -4 = 2190
2190 x 3 = 6570
To muluply by 9 just remember that 9 is one less than 10, so
all that is necessary is to add a zero and then subtract the original
multiplicand Tale 9 x 84
10 x 84 840
840 - 84 5756
This can be extended if asked to multiply by 18 all that is
necessary is to multiply by 9 and double the product For
crample, H8 18
48 x 10 = 480
4480 - 448 = 4032
4032 x 2 8064
Alternatively you can start from the fact that 18 is 20 less 2 in
which cause the sequence is
48 X 2 = 896
896 x 10 = 8960
8960 - 896 8064
This method can be used for all numbers which are multiples of 9
I or aarople usled to multiply 765 by St you will realize that
54 x equivalent 10 6 x 9 the calculation then gues
765 Y 64590
4570 10 4520)
47-4570 41310

·Multiplication 35
Multiplying by II is easily done if you remember that 11 is
10+1 Therefore to multiply any number by 11 all that is
necessary is to add a and then add on the original number
For example, to multiply 5342 by 11
5342 x 10 53420
53420 +- 5342 = 58762
In the case of 11 there is an even shorter method that can be
used if the multiplıcand is a three-digit number in which the
sum of the last two digits is not more than 9 For example if
asked to multiply 653 by 11 you check that 5 and 3 total less
than 9 and since they do you proceed as follows
First multiply the last two digits, 53, by 11
53 x 11 = 583
To this add the first, hundreds digit multiplied by 11
600 x 11 = 6600
6600 + 583 = 7183
Another case where this method can be used is with 121 Here
you have a choice, you can either work on the basis that 124 is
10 plus a quarter of 10, in which case 872 x 124
872 x 10 =8720
8720 – 4=2180
8720 +2180 = 10900
Alternatively, you can work from the basis that 124 is one-eighth
of 100, in which case 872 x 121

·36 Figuring 872 x 100 = 87200 87200 - 8 = 10900 It's best to check if the multiplicand is divisible by 8 before adopting the second way In fact this process of checking pretty well does the calculation for you If you have to multiply 168 by 12% a moments thought shows that 8 goes into 168 exactly 21 times All you then have to do is add two zeros to get the correct product - 2100 But if the multiplicand had been, say, 146 then clearly the first method is the one to use Here are some other relationships that can be exploited to use this basic method 1121 is 100 plus one-eighth of 100 125 is 100 plus one-quarter of 100, or 125 is one-eighth of 1000 45 is 50 minus 5, 50 is half of 100 and 5 is one-tenth of 50 25 is a quarter of 100 35 is 25 plus 10 99 is 100 minus 1 90 is 100 minus one-tenth of 100 and so on If you experiment you will find many more of these useful relationships all of which can be used to take advantage of the basic shortcut offered by the fact that to multiply by 10 all you do is add a zero After some practice you will find that you can spot almost without thinking a case where this first method is going to help METHOD II The next method I am going to describe can be used when the multiplier is a relatively small number, but one for which our ·Afultiplication 37 first method is unsuitable because there is no simple relationship to 10 which can be spotted and exploited We have in fact already made use of this second method without my drawing attention to it When, for instance, you multiply by 20 by the simple and obvious means of multiplying by 2 and then adding a 0, what you are doing is taking out the factors of 20, 2 and 10 and multiplying by them each in turn This method can be extended to any number which can be broken down into factors For example if you are asked to multiply a number by 32 you can break 32 down into its factors 8 and 4 and proceed as follows (here to start with anyway, I suggest the use of paper and pencil) To multiply 928 by 32 928 4 3712 8 29696 Even if you have to write the calculation out as abosc it is a great deal quicker than the conventional method, which would not only involve two sequences of multiplication but also one of addition In practice this method is best used where the multiplier is relacly small and its factors thercfure easily and quickly crtracted, if you remember the multiplication tables up to 12 you will be able to judge at a glance whether or not this is a suitable method in the case of a two-digit multiplier

·38 Figuring
METHOD III
This third method is one that should be used here there is no
casuy discernible means of using the first no, and where the
number of digits involved makes them impricicable
I am going to start by cxplaining the method with relatively
small numbers so that you can grasp the essentials
For our first example let us take 13 x 19 First add the unit
digit of one number to the other numbet, thus
9 + 13 = 22
You then think of that sum as so many tens in this case 22 tons
Now multiply the ISO unit digits together
3 x 9 = 27
Finally add the product to the tons figure you already have
27 + 22 (tens) = 247
Here is another example 17 x 14
4 + 17 = 21 (tens)
7 *4 = 28
28 + 21 (tens) = 238
There are particular short cuts for multiplying together to
igure numbers with the same tens or the same units figure
When the tcns figure is common you add to one number the
units figure of the other multiply this sum by the common tens
figure and add to this product considered as tens, the product
of the two units digts For example 49 x 42

·Multsplication 39
49 +2=51
51 x 4 = 204
Add to this product (2010, when considered as tens) the product
of the units (9 x 2 = 18) to get the final product of 2058
Another example 58 x 53
58 + 3 = 61
61 x 5= 305 (tens)
8 x 3 24
3050 + 24 = 3074
RE
If the common tens figure is 9 there is an even simpler method
Subtract each of the numbers from 100 Multiply the remainders
together - this gives you the last two figures of the final product
To arrive at the first two digits take away from one of the numbers
the figure by which the other was short of 100 For example,
93 x 96
100 - 93 = 7
100 - 96 = 4
4 x 7 = 28 your two last digits 93 – 4 or 96 - 7 gives you 89
your first two digits The product is therefore 8928
If the units figures of two two-figure numbers are the same
you can multiply them by adding the product of the two tens
figures (considered as hundreds) to the sum of the tens figures
multiplied by the common units digit and the square of the
common units digit For example 96 x 46
9 *4= 36 (hundreds)
9 + 4 = 13, 13 y 6 = 78 (tens)
6x6 = 36
3600 + 780 + 36 = 4416

·40 Figuring
Or to multiply 62 by 42
6 x 4 = 24 (hundreds)
6+4 = 10,10 x 2 = 20 (tens)
2 x 2 =4
2400 + 200 + 4 =
2601
If the common final digit is 5 it is even simpler - to the product
of the two tens figures considered as hundreds add half the sum
of the two tens figures still considered as hundreds Then add 25
to arrive at the final product For example 45 x 85
4 x 8 = 32 (hundreds)
4 + 8
= 6 (hundreds)
2
3200 + 600 = 3800
Add 25 to obtain the final product of 3825
If the sum of the tens is an odd number take it to the nearest
whole number below, and add 7) not 25, to obtain the final
product For example, 95 x 23
9 Y 2 = 18 (hundreds)
9 +2
2
=55 (hundreds)
1800 + 500 + 75 (25 plus 5 of 100)
2375
All of the methods I have described so far an be donc mentally
when you have had a little practise - I will now describe others
which can be used more generally but which require pencil and
paper Fren with these methods most of the calculations can be

·Multiplication 41 done mentally, you use the paper to keep note of your intermediate results in each case you do a sequence of diagonal or vertical multiplications, the pattern of these is shown in diagrams to the right of the examples For example, to multiply 63 by 48 write down the numbers thus 63 48 :1X1, 3x8=24 so put 4 in the units column and carry the 2, you will add this to the sum of the products of the 'diagonals' - 8 and 6, and 4 and 3 Your mental calculation runs thus 6x8 = 48, 4 x 3 ~ 12,48 + 12+2 - 62 You write 2 in the tens column and carry 6 - this you add to the product of the two tens digits - 6 and 4 The mental calculation 6 x 4 = 27, 24 + 6 = 30 gives the final figures - all you have had to write down the game.com Itself and the answer 63 48 3024 Set out three-digit figures to be multiplied togciher in the same way 436 254 1:XX X: 1::

·42 Figurmg
Again multiply the units digits and write down the units figur
of the answer 4, and carry the tens digit 2 Now multiply 3 by
and add the 2 you are carrying to make 14 Add this to the pro
duct of 5 x 6 = 30, to make a total of 44 You now have tre
figures of your final answer and are still carrying only one figure.
4 - in your head The figures you have written read
436
254
44
Your next mental steps are to add the 4 you are carrying to the
products of 4 x 4 6 x 2 and 3 x 5 and the calculation wil
run 4* 4 == 16, 16 + 4 = 20, 6 x 2 == 12, 12 + 20 = 32
3 x 5 = 15 15 + 32 = 47
Set down the 7 and carry the 4
Now multiply the left hand set of diagonals - 4 and 5, and
3 and 2 and add the carried 4 4 x 5 = 20 20 + 4 = 24
3x2=6,6 + 24 = 30
Set down the 0, carry the 3, and add it to the product of the
first hundreds digits 4 and 2 4 X 2=8 8 + 3 = 11
Now write 11 nert to the 0
436
254
110744
Again, all
you have had to write are problem and answer
If the multiplier has only cho figures you can still use this
method by replacing the missing hundreds figure with a 0 For
example set out 476 x 26 like this

·476
026
6 x 6 = 36, put down the 6, carry 3, and proceed as before
7x6=42, plus 3 carried 45, plus 2 x 6(- 12) = 57
Put down the 7 and carry the 5
4Y 6 = 24, plus the 5 carried = 29,6 x0=0 2 *7 = 14
29 + 14 = 43 Set down 3 and carry 4
4x2 = 8,8+ 4 = 12,7x0=0,12 + 0 = 12 - so you
write 2 and carry 1 The figures you have now written read
476
026
2376
4x0 = 0, but you are still carrying I so the final answer reads
12376
We can now extend the method to deal with four-figurc
numbers For instance to multiply 9246 by 2541 set down the
problem in the same way as before
2345 ::: 1 ::X :**
X: X:: 1:::
First the units 3 x6 =18, write down the 8 carry the I Next
the first pair of dugonals 4x3 = 12, 12+1 = 13, 4x6 =
24,24 + 13 = 37

·44 Figuring Write down the 7 and carry the 3 9246 2543 78 Now 2 x 3, 4 x 4, and 5 x 6 2 X 3 = 6 5 x6 =30 4 x 4 = 16 6 + 3 (carried) 9 30 + 9 = 39 16 + 39 = 55 Ser down 5 and carry 5 9246 2543 578 Now the four sets of diagonals 9 X 3 =27 2 x6 =12 2 x 4 = 8 5 X 4 = 20 27 + 5 (carried) = 32 12 + 32 44 8 +44= 20+52 72 52 Set down 2 and carry 7 9246 2543 2578

·Multsplication 45
By the same procedures as before
9 X 4 = 36
4 x 2 = 8
2 x 5 = 10
36 + 7 (carried) = 43
43 + 8 = 51
51 + 10 = 61
gives us another digit of the solution, and 6 to carry
9x5= 45
2 x 2-4
45 + 6 (carried) = 51
51 +4=55
gives another, 5, with 5 to carry
9246
2543
512578
and we get the final two digits by multiplying 9 x 2 and adding
the 5 we are carrying 18 + 5 =23
9246
2543
23512578
This same method of adding the products of groups of diagonal
multiplications can be used for larger numbers, but a little more
paperhorl will make it simpler For instance to multiply 637432
by 513124 start by setting the numbers out in the usual way
637432
513124

·46 Figuring then proceed as follows 45 (4 x 2) (4 x 3) + (2 x 2)= 12+4= 16 (4 x 4) + (2 1) +(3 x 2) = 16+2+6= 24 (4 x 7) + (2 X 3)+(3 x 1) +(4x2)= 28 +-6+3+8= (4 x 3) + (2x1)+(2 x 7) + (3 x 3)+(4x 1) = 12+ 2 + 14 +9+ 4 = 41 (4 x 6)+(5 x 2)+(2 x 3)+(1 x 3)+(7 x 1) + (3 x 4)=24+10+6+3+7+12= 62 (2 x 6) +(5 x 3) +(1 x 4) +(3 x 1)+(7 x 3) = 12+15+4+3 +21= 55 () x6)+(5 x 4)+(1x 7) + (3 x 3) = 6 +20 +7+9= 42 (3 x 6) +-(5 x 7) + (3 x 1)= 18+35+3= 56 ( 16)+(5 x 3)=6+15= 21 (6 x 5)= 30 327081657568 All the computing goes on in your head the addition at the end involves only two digits in each column and the amount of paperwork is limited to the figures on the right-hand side of the page CHECKING BY CASTING OUT THE NINES No one is immune from error, and quick ways of checking long computations are always useful The method I describe here is not infallible but if the answer you have obtained checks out

·Multiplication 47
as correct when you have used it, the chances of your being wrong
are very slım indeed This is the way it is done
Add up the digits in each factor (the numbers you are
multiplying together) and in the product
Divide each of the sums of digits by 9
Set down the remainder in each case - this is known as the
check number
Multiply the check number of the multiplicand by the check
number of the multiplier, and add up the digits in the product
If this gives you the same number as the check number of the
original product you can assume that that product was correct
For example to check that 8216 x 4215 = 34630440
8+2 +1 +6=17, casting out the nines leaves 8 as check
number
4+2 +1+5=12, casting out the nines leaves 3 as check
number
3+ 4+6 +3+0+4 + 4 +0=24, casting out the nines
leaves 6 as check number
3x8=24, 2+4= 6, which was the check number of the
onginal product

·50 Figuring Add each column of figures separately Put the units figure of the total directly below the column you have just added, the tens figure one line down and one place to the left Repeat this process for each column and add the two rows of figures together to obtain the final result Another method of avoiding carrying also involves adding each column separately The column totals are set out in a staggered line, the units figure of the second column below the tens figure of the first the units figure of the third column below the tens figure of the second and so on The column totals are then added to give the final answer Here is an example 962853 524861 212346 401258 864321 19 22 24 13 15 28 2965639 (You can of course work from left to right if you wish and set each column total ore place further right rather than one place further left) 962853 524861 212346 401258 864321 28 15 13 24 22 19 2965639 ·Addition 51 There arc cen tricks you can play with addition Ask a friend to write down any fire-figure number say he writes 21564 You apparently at random, choosc figures to stitc bclox his You put down 78435 - these are in fact the digits which added to those they stand below, will total 9 You now ash your friend to add a further five-figure number, and you again if rite below his digits those that would make them up to 9 Your friend adds a final fifth line of five figures and you instantly draw a line and add all five numbers together - the total will always be the last number your friend #Tote with 2 subtracted from the last digre and 2 inserted before the first one 21564 784354 12567 87435 4245 242143 CHECKING BY CASTING OUT THE NINES Additions, like multiplications can be checked (though not infallibly checked) by casting out the nines First add up the digits of cach number in the sum, and of the answer you had arrived at Divide each of these numbers by 9 and set down the remainder Total the remainders of the numbers you were adding together, and again cast out the nines If the remainder you are left with is the same as the remainder when the sum of the digits of your answer was divided by 9 jou may assume that answer was correct.

·5
DIVISION
Division is the reverse of multiplication - just as subtraction is
the reverse of addition To multiply 3 by 4 is to add four threes
together and find the total - 12 Dividing 12 by 3 could be said
to be subtracting threes until you had nothing left - and you
would of course do it four times But division seems more
intractable to most people, and particularly difficult to handle
mentally In this chapter I will show how division by particular
numbers can be handled in particular ways, and, at the end of the
chapter, how you can check to see whether a number is or is not
divisible by any number from 2 to 11 (or, of course any multiple
of such a number) But first some special cases
To divide a number by 5 you take advantage of the fact
that 5 is half of 10 multiply the dividend (the number being
divided) by 2 and divide by 10 by moving the decimal point one
place to the left For example
330
165 divided by 5 =
10
To by 15 multiply the dividend by 2 and divide by
30 For example
210
105 divided by 15 = 7
30
<33
52

·Decuriore 53 If you simplify in this way by doubling divisor and dividend - and are left at the end of the calculation with a remainder you must remember that this, 100, will be doubled Some divisions an be siraplified by halving both divisor and dividend - divisions by 14, 16, 18, 20, 22, and 24 become divisions by 7, 8, 9, 10, 11, and 12 - but this time you must remember to double the remanaer Here are some examples 392 dinded by 14 196 7 = 28 464 divided by 16 232 8 29 441 882 divided by 18 - =49 9 2480 4960 divided by 20 = 10 248 473 946 divided by 22 43 588 1176 divided by 24 = DIVIDING BY FACTORS If a number can be broken down into factors it may be simples co divide by these successively, than to do a single calculation A mental division by 8, and then by 4, as simpler than a division by 32 For example to divide 1088 by 32 first divide by ·54 Figuring 1088 8 = 136 and then divide the answer, 136, by 4 to get the final result – 34 Numbers casy to handle in this way are the products in the basic multiplication tables multiples of 11 for instance go particularly smoothly - To divide 2695 by 55, first divide by 5 2695 5 539, and then by 11 539 11 49 to get the answer To divide by numbers that are powers of 2 (4,8 16 and so on), you merely have to go on halving the dividend 16 for instance is 24 so halving the dividend four times is the same as dividing by 16 To divide 8192 by 16 halve once to get 4096 and again to get 2048 and a third time to get 1024 and finally a fourth time to get the answer 512 This technique makes dividing by high powers of 2 casy - for instance to divide 32768 by 128 - which is 27

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instance you are disading by if it is simple neuters
numbers in the calculation by 4 - desine tutti
onginal dividend by 30 For example to da fritt

·56 Figuring
360 X 4 = 1440
71 x 4= 30
1440
= 48
30
By the same principle when dividing by 12) multiply the dividend
by 8 and dinide by 100, when dividing by 371 multiply by 8 and
divide by 300 and when dividing by 621 multiply by 8 and divide
by 500
To divide by 14 double the dividend and the divisor and divide
by 3 (but remember to halve any remainder)
To divide a number by 24 double the dividend and the divisor
and divide by 5 - and go about dividing by 3} in the same way
CHECKING WHETHER A NUMBER IS
EXACTLY DIVISIBLE
There are tests that can be made to show whether a number is
exactly divisible by another number (or multiple of t) Here are
some of therm
If a number is divisible by two it will end in an even number
or a 0
If a number is divisible by three the sum of its digits will be
divisible by 3 (for erample 372 = 3+2+7= 12) A corollary
of this is that any number made by rearranging the digits of a
number divisible by 3 will also be divisible by 3
If a number is divisible by four the last two digits are divisible
by 4 (or are zeros)

·Dro son 57
If a number is divisible by five the last digit will be 5
or 0
If a number is divisible by suy the last digit pill be even and the
sum of the digiis divisible by 3
There is no quick test for finding out of a number is divisible by
seyen
If a number is divisible by eight the last three digits are divisible
by 8
If a number is divisible by nine the sum of its digits is divisible
by 9
If a number is divisible by ten it ends with a 0
If a number is divisible by eleven the difference between the
sum of the digits in the even places and the sum of the digits in
the odd places is 11 or 0 For example, 58432 is shown to be
divisible by 11 because
5+4 + 2 = 11
8 + 3 = 11
11 - II = 0
25806 is shown to be divisible by 11 because
2 + 8 +6 = 16
5 +0 5
16 -5 -11

·58 Figurmg CHECKING BY CASTING OUT THE NINES - Divisions like additions and multiplications, can be checked by casting out the nines although, again, the method is not infallible You obtain check numbers, as before, by adding up the digits in the numbers in the calculation - in this caso the divisor, dividend and quotient - and dividing them by 9 The check number is the remainder left after each division Multiply the check number of the devisor by the check number of the quotient If the check number of this p oduct is the same as the check number of the dividend the division may be assumed to be correct For example to check 2426376 5321 = 456 2+4+2+6 +3+7+6= 30, casting out the nincs leaves 3 5 + 3+2+1=1l, casting out the nines leares 2 4+ 5+-6 == 15 casting out the nincs leaves 6 6 x 2 = 12 (check number divisor x check number quotient) Casting out the funes leaves 3 - which is the same as the check number of the dividend When there is a remainder the sum of the digits in the remainder are added to the product of the check numbers of the quotient and the divisor and the operation is completed in the usual way for example to check 1481265 = 346 remainder 39 4281 1+4+8+1+2+6+5=27, casting out the nines leaves 0 4 +2+8+1 = 15 casting out the nincs leaves 6

·Driston 59
3+4 +6= 13, casting out the nines leaves 4
3+9 = 12, casting out the nines leaves 3
6 x 4 = 24, 24 +3 = 27, adding these digits and casting out
the nines leaves 0 - which is the check number of the dividend

·6 SUBTRACTION Subtraction is the opposite of addition, but many people find adding easier If the number being subtracted is a multiple of 10 however the sum is easy, and you can simplify subtractions by adding to both the number being subtracted, and the number being subtracted from the figure which will bring the former up to round figures It is a particularly useful trick when subtracting sums of money in decimal currencies For example to subtract 46 from 58 add 4 to both numbers - the problem then becomes the easy one of 62 - 50 = 12 Or to deduct $1 72 from $3 64, add 8 cents to each sum $372 – $1 80 = $1 92 When you are dealing with three-figure numbers you will bring the number being subtracted up to the nearest hundred 246 - 182 becomes 264 - 200 and the answer, 64 obvious And with four-figure numbers you add to both numbers the number which will bring the number being subtracted up to the nearest thousand When 2348 - 1821 becomes 2527 - 2000, there is no problem in seeing that 527 is the answer 60 ·Subtraction 61 CHECKING ERRORS IN SUBTRACTION The simplest check is to add your answer to the number being subtracted - the sum will be the same as the number being subtracted from, if you have made no mistakes You can also use the method of casting out the nines, which I have described already - in this case the procedure is to add up the digits of each number in the subtraction, dide by 9, and note the check numbers - the remainders - arred at Having cast out the nines in this way subtract the check number of the number being sub tracted, from the check number of the number being subtracted from, if the figure arrived at is the same as the check sumber of your answer you can assume that your original subtraction was carried out correctly But again remember that masuing out the nincs is not infallible

·7
GCM and LCM
I have already explained how numbers can be broken down into
factors - numbers which multiplied together, give the original
number as a product
The Greatest Common Measure, or Highest Common Factor,
is the greatest number that is a factor common to two numbers -
it can be defined as the greatest number that is contained an
exaut number of times in each of two or more numbers and
abbreviated to GCM or HCF
The usual way of finding the GCM of two numbers is to break
both of them down into their factors, pick out the factor or
factors they have in common, and multiply these together
For instance to find the GCM of 72 and 126
72 = 2 X 2 X 2 X 3 X 3
126 = 7X2 X3 X 3
The common factors are 2 3 and 3 and the GCM, the product
of these factors multiplied together 18
Or to find the GCM of 7293 and 442
62

·GCA1 and LCM 63
7293 = 3 x 11 x 17 x 13
442 = 2 x 13 x 17
The GCM is therefore 13 x 17, or 221
As larger numbers are harder to split into factors another method
can be uscá This involves dividing the larger number by the
smaller, and then dividing the smaller number by the remainder
from the first division The remainder of the first division is then
divided itself by the reminder from the second and so on until
no remainder is left The last divisor is the GC\I of the two
numbers you started with Here are two examples
Find the GCM of 4781 and 6147
4781)6147(1
4781
1366)4781(3
4098
683)1366(2
1366
The required GCM s 683

·64 Figuring Find the GCM of 13536 and 23148 13536)23148(1 13536 9612)13536(1 9612 3924)961262 7848 1764)3924(2 3528 396)1764(4 1584 180)396(2 360 36)180(5 180 The required GCM is 36 LEAST COMMON MULTIPLE The Least Common Multiple is the smallest number which contains an exact number of each of two or more numbers The usual method of finding the Least Common Multiple (LCM) of two or more numbers is to break them down into their prime factors - factors which cannot themselves be split into smaller factors

·GCM and LCM 65
From the prime factors of all the numbers choose the highest
powers of each factor that occurs Multiply these together, and
the product will be the required LCM
For example, to find the LCM of 78, 84, and 90
78 = 2 x 13 x 3
84 = 2 X 2 X 7 X3=2' x 7 x 3
90 2x3 x3 x 5= 2 x 32 x 5
Therefore the LCM 18 22 x 32 x 13 x 7 x 5 = 16380
To find the prime factors set it out in this way (the problem is to
find the prime factors and LCM of 192, 204, 272)
2)192, 204, 272
2) 96, 102, 136
2) 48,
51,
68
2) 24,
51, 34
3) 12,
51,
17
17) 4,
17,
17
4,
Each number is divided by factors common to at least two of
them when no more divisions are possible, multiply together
all the numbers you have disided by and the numbers in the
bottom ron, to obtain the least common multiple

·66 Figuring
2 X 2 X 2 X 2 X 3 X 17 X 4 = 3264
When the numbers involved are very large it will be quicker to
use the GCM of the numbers as the first divisor
Find the LCM of 7535 and 11645
The GCM of 7535 and 11645 is found to be 685 Dividing by
the GCM, we obtain
685)7535, 11645
11
17
Therefore the LCM 15
685 x 11 x 17 = 128095
Find the LCM of 4781 and 6147
The GCM of 4781 and 6147 is found to be 683
4781 x 6147
683
43029
The lowest common multiple of two numbers is the same as the
product of the two numbers divided by their highest common
factor (Conversely the GCM and the LCM multiplied together
gives the product of the two numbers )
When the LCM of a list of numbers is wanted find the LCM
of the first (wo numbers, then the LCM of the third number and
the LCM of the first two and so on
Find the LCMI of 385 231 165 and 105

·GCM and LCM 67
The GCM of 385 and 231 is 77 Therefore the LCM of 385
and 231 is
385 x 231
77
1155
The GCM of 165 and 1155 is 165 Therefore the LCM of 165
and 1155 is
165 x 1155
165
= 1155
Now the GCM of 1155 and 105 15 105 Therefore the LCM of
1155 and 105 18
1155 X 105
105
= 1155

·8
SQUARES and SQUARE
ROOTS
000
000
The square of a number is that number, multiplied by itself it
is also - and this explains why it is called a square - the number
of units in a square grid having sides of a given number of units
Think of building up squares by laying down marbles - the
first square will have one o the second the third ooo and
so on One, four, and nine are thus the squares
So 4 squared is 4 x 4 It is written with a figure * (44) and is
known as the second power of four Four Squared equals 16,
and 4 is therefore the square root of 16 - which is written
thus V16
The sequence of squares and the sequence of odd numbers
have what at first sight seems a mysterious relationship
1 = 1=1
1+3 = 422
1+3+5= 9=3*
1+3+5+ 7 = 16 = 45
1+3+5+7+9= 25 = 5*
1 + 3 + 5 + 7 + 9 + 11 = 36 = 62
1+3+5+7+9+ 11 + 13 =4971
And so on
63

·Squares and Square Roots 69
But if you think of the square of marbles the reason for it becomes
clear
1 2 3 4 5 6 7 8 9
To make up the second square one must add 3 marbles to the
1 square To make up the third square one must add 5, and so on
This diagram also makes it clear why another relationship
holds any triangle number added to the next highest triangle
number always gives a square number (Triangle numbers as
their name suggests, are those that can be made from units in a
triangular display 36, 10 and so on)
OOO
000
The triangular numbers have another special relationship to
the square numbers multiply any triangle number by 8 and add 1
and
you
have a square number The sequence runs

·70 Figuring Tringle 11 (8 x 1) + 1 =9 Triangle 2 = 3 (8 x 3) +1=25 = 5! Triangle 3 =6 (8 x 6) +1=49 = 78 Triangle 4 =10 (8 x 10)+1=81 = 9+ Triangle 5 = 15 (8 x 15) + 1 = 121 = 11% Note that the odd sequence of numbers 3, 5, 7, 9, 11, etc shows up once again Mathematicians have been fascinated by the connections between square and triangle numbers, and between ordinary numbers and numbers of special shape for centures They are still trying to find new connections between them Numbers ending in 5 can be squared mentally very easily ignoring the 5 muiuply the figures left by the number which follows it numerically, insert 25 after the product of this mulaplication and you have the square of the original number For cxample 651-67= 42, with 25 at the end, 4225, and 145º = 14 x 15, which is 210, with 25 on the end, 21025 Numbers made up only of threes have a special pattern of squares 33! = 1089 333 = 110889 3333 = 11108889 33333 = 1111088889 333333! = 111110888889 and so on Squares for other numbers can be found quickly by using the methods described in the chapter on multiplication ·Squares and Square Roots 71 FINDING SQUARE ROOTS This is the conventional method of finding square roots First divide the number up into groups by putting a mark over every second figure - starting with the units digit, going on to the hundreds digit and so on For instance if the problem is to find the square root of 233289 set it out like this 233289 Now find the largest number the square of which is less than 23 233289(483 16 88)732 704 963)2889 2889 this is 4 - which gives you the first figure of the answer Insert this to the right of the number and subtract 42 (16) from 23 Now bring down the next group of figures – 32- and put it after the remainder just obtained to make 732 The first part of the divisor of 732 (obtained by doubling 4, that part of the root you have already obtained) is 8 The final digit of the divisor, and the second of the root itself, is arrived at by trial It will be the largest number that, inserted after the 8, and then used to multiply the number thus obtained will give a product of less than 732 It turns out in this case to be 8 - for 8 x 88 704 and 732 - 701 = 28 ·77 847 Now go through the same process again Bring down the next group of numbers, 89, to make 2889 Arrive at a divisor by doubling 48 (the part of the root already arrived at) This gives you 96 Find by trial the next number of the square root (in this case the final one), which will also be the last digit of the divisor It turns out to be 3, and you now have the complete root, 483 which can be checked by multiplying it by itself Square roots by factors Square roots can also be found by breaking numbers down into factors which are the squares of known numbers (4 9 16 25 and so on) Multiplying together the square roots of the factors will then give the square root of the number For instance the number 20736 is found to have the factors 4 y 4 x 4 x 4 x 9 x 9 The square root is then 2 x 2 x 2 X 2 x 3 y 3 which equals 144 B th these methods are time consuming howcver - and there are shortcuts which, given a little practice allow you to arrive at squarc roots mentally Tirst you must learn the table of squares of numbers from 1 to 9 by heart number square 1 2 3 4 5 6 7 8 9 9 16 25 36 49 64 81 ·Squares and Square Roots 73 You will see from this table that any square ending in I will have a root ending in 1 or 9 A square ending in 4 Mill have a root ending in 2 or 8 A square ending in 9 will have a root ending in 3 or 7 A square ending in 6 a root ending in 4 or 6 and a square ending in 5 a root ending in 5 So any number that is a perfect quare will end in 1, 4, 9, 6, or 5, and, with the exception of 5, the final number of the square will indicate two possible values for the last digt of the square root This table is the basis of a quick method of finding square roots To find the root of a four-figure number first break the number into two groups of two digns For example 6241 breaks into 62 and 41 Now consider the group with the thousands and the hundreds digits We know that the square root we are looking for will have two digits - to find the first we think where 62 stands in relation to the memorised table of squares It is less than 64 but more than 49, so the highest possible first digit of the root is 7 We must now find the units digit of the squarc root We know from our table that it must be 9 or 1 - for only they give squares ending in 1 To decide which take the number already arrived at as thc first number of the root - in this case 7, multiply it by itself plus 1 If the product is more than the first two figures of the number for which you are finding the root take the lower of the two possible digits figures - in this case 1, to arrive at the square root if it is less take the higher of the two possible figures -7% (7 of 1)-7X 856, which is less than 62 The square root of 6241 is thercfore 79 Herc is another example ſo find the square root of 4096 40 lies between 36 and 49 so the first figure of the square root is 6 The second figure could be 4 or 6 6 7 7 = 42, which is more than 40 so we take the lower of the possible units digits The square root is therefore 64 When the square has five or six figures we know the square ·74 Figuring root will have three To find the square foots of numbers up to 40000 you must remember more squares - those up to 20 number 11 12 13 14 15 16 17 18 19 20 square 121 144 169 196 225 256 289 324 361 400 With this table memorised you can set about finding the square root of longer numbers which are perfect squares Let us take an example To find the square root of 15129 First divide the number into two groups of digits by taking out the last two figures This gives 151 29 From the table you know that 151 falls between the squares of 12 and 13 - which are 144 and 169 The first ewo figures of the root you are extracting will therefore be 12 - the square root of the lower figure To find the final figure you use the method alrcady described the number from which you are extracting the square root ends in a 9 the digit you are looking for must therefore be a 3 or a 7 To find which add 1 to 12 (the first figures of the root which you have already arrived at) and multiply the sum by 12 The product 156 15 gtcater than 151 so you take the ·Squares and Square Roots 75 lower of the two possible digits You now have the complete square root of 15129, it is 123 Here is another example To find the square root of 30276 Divide the number into two groups of digits as before 302 76 From the table we know that 302 falls between the squares of 17 and 18 The first two digits of the squarc root are therefore 17 The number ends in 6 so the final figure must be 4 or 6 17 x 18 = 306 this is greater than 302, so you take the lower of the possiblc digits to arrive at the square root of 174 If the number from which the square root is to be extracted is higher than 40000 you use a slightly different method For example, to extract the square rout of 537289 First break it up into groups of two figures, starting from the right 53 72 89 To find the hundreds figure of the square root go to the first memorised table 53 stands between the squares of 7 and 8 The lower of these numbers is our first figure, 7 To find the tens figure we must find the difference between 53 and the square of 7, that is between 53 and 49 53 -- 494 We now put 4 before the left-hand digit of the second group of figures in the number from which we are extracting the square This gives 47 We divide this number by twice the figure already arrived at as the first in the square root - 7 - plus 1 This gives us 15 ·76 Figuring 47 15 -3, with remainder 2 (note that you are obtaining the quotient which will gnc the least remainder abone or below - the remainder is the number that must be added to or subtracted from the dividend to bring it to the nearest multiple of thc divisor) The quotient thus obtained is the tons digit of the square root we are extracting, so the first two diguts of the root are 73 We know that the units digit must be 7 or 3, for the number ends in a 9 As thc quotient obtained, 3, was greater than the remainder 2, we take the smaller of the eno possible figures and arrive at the complete square root of 733 When the quotient is smaller than the remainder take the larger of the two possible figures as the units digit ·9 CUBES and CUBE ROOTS You will remember that square numbers could be illustrated by rectangular arrays of marbles Cubrs can be illustrated with square building blocks lou build a cube by adding layers of blocks to a squarc until there arc as many layers of blocks as there are blocks in one side of the square A square made up of only i block is also block deep A square with a side of 2 blocks long will be made into a cube by adding another layer - there are 4 blocks in each layer and 8 in the whole cube 2 x 2 x 2 = 8 A square with sides 3 blocks long contains 9 blocks, and a cube with sides of 3 blocks three times as many 3 x 3 x 3 =27 Going on up, enlarging the cube by onc unit each aime, ac find the fourth cube contains 64 blocks the fifth 125, and so on A cubic number is onc obtained by multiplying a whole number by itself and then by itself again The cube root is the number - the whole number you started with Cubes

·78 Figuring
said earlier, written with a superior 3 So 64 can be written as 4:
4 on the other hand can also be written as V64, that is to say
the cube root of 64
Another way of indicating the cube root of a number is by
what is called a fractional index
641
or, in words 64 raised in the power of l over 3
The cubic numbers like the square numbers have a special
relation to the odd numbers
1% = 1
23 = 8 3+5
3= 27 = 7+9+11
48 = 64 = 13 + 15 + 17 + 19
58 = 125 = 21 +23+25+27 +29
68 = 216 = 31 +33 +35 - 37 +39 +41
75 = 343 =43 +45 +47 +49 +51 +53 + 55
and so on
And cubic numbers, like square numbers have a special
relationship to triangular numbers it is an odd one If the cubic
numbers are added up in order (1', 1+23 18 +2+3*,
and on the sur arrived at are the squares of the sequence of
triangular numbers

·Cubes and Cube Roots 79
1% =
15 19
1 = iriangle number 1
13 + 2 = 9= 32
3 mangle number 2
1' + 24 + 38 = 36 = 62
triangle number 3
1 + 23 + 39 43= 100 = 102
10 = triangle number 4
19 +2° +33 + 40 + 53 = 225 = 152
15 = triangle number 5
1s +2° +34 +45 +53 +63 441 = 21
21 = Inangle number 6
10+23 + 38 +48 +58 +68+73 787 = 282
iruungle number 7
18+23 + 3% + 48 6* + 7 + 8 = 1296 = 362
36 = Inungle number 8
28 =
and so on
EXTRACTING CUBE ROOTS
The most common conventional way of extracting cubc roots
is by factors - here, for exafnple, is how you extract the cube root
of 474552 by this method

·80 Figuring
2)474552
2)237276
2)118638
3)59319
3)19773
3)6591
13)2197
13)169
13)13
1
The figure 474552 may thus be expressed as
23 x 33 x 138
3,474552
323 x 33 133 - 2 x 3 x 13 = 78
But there are simpler methods for extracting the roots of numbers
which are exact cubes - I will show first how it can be done for
numbers of up to sex digits
Again there is a table to remember this time of the cubes of
the numbers from 1 to 10

·Cubes and Cube Roots 81 number rubes 1 1 2 3 4 5 6 7 8 9 10 27 64 125 216 343 512 729 IX The hrst thing to notice about this table is that the numbers from 0 to 9 all occur If a number is cubic one can therefore determine the units figure of the cube root immediately If a cubic number ends in 1,4 5 6 9 or 0 its cube root will end in the same number If it ends in 2 the cube root will end in 8, and if it ends in the cube root will end in 2 Similarly if the number ends sn 7 the unus digte of the cube root will be 3 and if it ends in 3 the units digit of the cube root will be 7 You will also need this table when you come to determine the tons digit If a cube has four, five, or six digits the root will contain two To extract the root fir divide the number into two groups by counting back three places from the units digit For instance if you are extracting the cube root of 50653 dude it thus 50 653 Tend the first digte of th- cube root by referring back mentalla to the table of cubes 50 falls between 27 and 64 - that is 32 and 43 the tens figure of the cube root you are be the root of the loner number - in thus ase 3 ·82 Figuring Now you must find the units digit, and you know from the table that as the cubic number ends in 3 its cube root will end in 7 The cube root you are seeking is therefore 37 Here is another example To find the cube root of 681472 First divide the number into groups 681 472 681 falls between the cubes of 8 and 9 The tens unit is therefore 8 The whole number ends in 2 the cube root therefore ends in 8 The cube root is 88 To extract the cube root of longer cubic numbers - those with seven eight or nine digits - is almost as simple once you have learnt yet another table 1 X 1 2 3 4 5 6 7 8 9 10 7 9 5 3 8 6 2 4 10 A number with seven cight or nine digits will have a cube root with three digits To find the root we again begin by dividing the cubic number into groups Moving from night to left male 2 ·Cubes and Cube Roots 83 break every three digits For example to find the cube root of 92345408 first divide it thus 92 345 408 To find the hundreds unit of the cube root think back to the table of cubes from 1 to 10 92 falls between the cubes of 4 and 5, 50 4 is the hundreds unit of the root we are extracting The units digit- the terminal figure of the cube root - must, we know, be 2, for the last digit of the cubic number is 8 Now we must find the tens digit To do this first add up the odd digits of the original number - the first third, fifth, seventh and (if it is there) ninth figures in the number, reading from right to left Subtract from this number the sum of the even figures in the number (the second, fourth, sixth and eighth digits) added up in the same way If the total of the even figures is greater than that of the odd figures add 11 to the latter before subtracting In the example this gives us 8+4+4+2 = 18 (the sum of the odd digits) from which we subtract 0+5+3+9 = 17 (the sum of the even digits) to get a remainder of 1 Now refer back to the table with columns and ) find in the X column the number just obtained as a remainder and notc the number opposite it in thc Y column Add together the cho digits of the final root which you already have (the hundreds and the units digits) and subtract from the sum the figure you have just obtained from the 3 column of the table

·84 Figuring
In the example the remainder was 1, this stands opposite 1 in
the table, so we subtract that figure from the sum of 4 and 2
4+2 = 6
6-1= 5
which gives us the tens unit of the cube root The complete
root is 452
If the sum of the hundreds and units digits is less than the
figure obtained from column Y of the table add 11 to that sum
before making the subtraction Here is another example
Find the cube root of 665338617
Divide the number into groups
665 338 617
The hundreds digit of the root is 8 as 665 stands between the
cubes of 8 and 9
The units digit is 3 as the number ends in 7
The sum of the digits in the odd places is
7+6+3 +5 +6=27
from which is subtracted the sum of the digits in the even places
27 - (1 + 8 +3+6) = 27 - 18 = 9
In the table 9 in the X column stands opposite 4 in the Y
column The sum of the digits already arrived at (8 + 3 =
minus the digit from the Y column, gives us the tens unit of the
cube root (11 – 4= 7)
The cube root is therefore 873
11)

·Cubes and Cube Roots 85 EXTRACTING FIFTH ROOTS When a number has been raised to the fifth power the root is surprisingly easy to find If the fifth power has between six and ten digits the root will have two The units digit of all fifth roots is the same as the units digit of the fifth power of the number, so all you have to find is the tens digit To do this divide the number at the fifth place, reading from right to left For example if the number you are finding the fifth root of us 33554432 divide it thus 335 54432 For the next step you need another table - that of fifth powers of numbers from 1 to 10 number fifth power 1 1 32 243 1024 3125 2 3 4 5 6 7 8 9 10 7776 16807 32768 59049 100000 When you have committed this to memory you can quickly determine where the left-hand group fits in the table In the example 335 comes between 35 and 46, the fifth root of 33554432 is therefore 32 here as with square and cube roots the root of ·86 Figuring the lower of the pair of powers that bracket the value of the first group in the table gives you the first figure of the root you are secking Here is another example Find the fifth root of 1073741824 First divide at the fifth place 10737 41824 10737 comes between the fiſth powers of 6 and 7 in the table the fifth root is therefore 6 followed by the final digi of the original number 64 By similar methods other roots can be found Ninth, thirteenth Seventeenth twenty-first and twenty-fifth roots are like fifth roots in that the power and the root have the same terminal digityou can extract the root therefore by constructing and memorising tables of powers from 1 to 10 Other roots - fourth sixth and so on can be extracted by the same methods as those described here for square cube, and fifth roots try working some of these out for yourself ·10 PERCENTAGES, DISCOUNTS and INTEREST Percentages -- fractions expressed as parts per 100 - are particularly useful in money sums rates of interest and discounts are both calculated in this way There are a number of shortcuts that can be used to simplify percentage sums and make them easier to do mentally For instance the problem may be to discount $400 by 30 per ant The normal calculation runs as follows 400 x 30 100 = 120 Deduct 120 (30 per cent of 400) from 400 to arrive at the answer 400 - 120 = $280 A quicker sechod, that can be done mentally is to deduct 30 per cent from 100 - which gives you 70 70 per cent of $100 is $70 so 70 per cent of $400 is four times as much - $280 Even if you are not working with round figures, and the amount to be discounted has cents as well as whole dollars face pence as well as pounds or paise as well as rupees - the calculations are of course the same for ny detimal ·88 Figuring the method of arriving directly at the discounted price, rather than finding the amount of the discount and deducting it, works For instance to find the discounted price of a sewing machine that is offered at a discount of 30 per cent on the marked price of $185 60 you must calculate 70 per cent of that sum This involves multiplying by 70 and dividing by 100 - or, (having divided both by 10) multiplying by 7 and dividing by 10 To divide by 10 shift the decimal point one place to the left, you can then arrive at the answer by a simple multiplication by 7 18 56 x 7= 129 92 which gives the answer of $129 92 directly Similarly to find $496 20 less 20 per cent, multiply 49 62 by 8 to arrive at the answer of $396 96 To find what remains when a sum is discounted by 10 per cent move the decimal point one place to the left in the original amount and subtract the amount thus arrived at from the original sum for example to discount $28 50 by 10 per cent deduct $2 85 to arrive at the discounted price of $25 65 A 90 per cent discount leaves 10 per cent of the original price - a simple shift of the decimal point one place to the left 51vcs the answer When more complicated percentages are being worked out it is often simpler to break the figures down For instance to find 32) per cent add 25 per cent to 5 per cent to 21 per cent By this method you would arrive at 32 per cent of $280 in the following way ·Percentages Discounts and Interest 89 25 per cent of 280 = 70 10 per cent = 28 so 5 per cent of 280 14 24 per cent of 280 (half of 5 7 per cent) 32 per cent of 280 = 91 Simularly to calculate 674 per cent of $460 230 50 per cent of 460 10 per cent of 460 5 per cent of 460 (half of 10 per cent) 21 per cent of 460 (half of 5 per cent) 46 23 11 50 674 per cent of 460 = 310 50 The break down can be done in various ways - the sum above, for instance, could be done like this 50 per cent of 460 121 per cent of 460 (quarter of 50 per cent) 5 per cent of 460 (tenth of 50 per cent) 230 57 50 23 - 671 per cent of 460 = 310 50 Here are two more examples - if the fractions are carefully chosen quite complicated percentages turn into simple additions of parts, or parts of parts 25 per cent of 298 124 per cent of 298 (half of 25 per cent) 5 per cent of 298 (fifth of 25 per cent) 74 50 37 25 14 90 424 per cent of 298 126 65 ·Jo Figuring To find 521 per cent of $368 50 per cent of 368 21 per cent of 368 (twentieth of 50 per cent) 184 9 20 521 per cent of 368 * 193 20 The general principle can be applied 3hy percentage sum – you will find ways of doing most of them in your head Bu mustakes in money sums are likely to be costly so check the resul by reversing the calculation - having found what 421 per cent of a given sum is find what sum your answer would be 42) per cent of SIMPLE INTEREST Interest paid on a loan or a deposit is also expressed as a per centage This states the amount due each year - the amount to be paid by the borrower to the lender When less than a whole scar is involved the amount due is a fraction of the yearly sum Ordinary interest is arnved at by assuming there are 360 days in the year and writing the fraction as the number of days for which interest is due over 360 Exact inte est is determined on a 365-day year, and is used where large sums are involved and generally by banks and financial institucions The usual way of doing interest sums is this multiply the number of days of the interest period over 360 or 365 by the amount of the loan by the interest rate over 100 For instance the ordinary interest on $326 for a period of 60 days at 6 per cent would be worked out as follows 60 X 326 x 6 360 x 100

·Percentages Discounts and Interest gr
Cancelling out simplifies the sum to
326
100
=326
This is the interest due
The 6 per cent and 60 days, and the 360 days of the ordinary
interest year cancel out nestly This fact can, in less obvious
interest calculations, help you to arrive at methods of doing them
mentally For instance to find the ordinary interest on $398 for
70 days at 6 per cent (the calculations are to nearest whole
penny)
= 398
Interest for 60 days at 6 per cent
Interest for 10 days (sixth of 60 days) = 66
Interest for 70 days
- 4 64
Or to find the ordinary interest on $426 for 36 days at 6 per
cent
Interest for 30 days (half of 60 days) = 2 13
Interest for 6 days (tenth of 60 days) 42
Interest for 36 days
2 55
Knowing that ordinary interest of 6 per cent for 60 days is
always one-hundredth of the principal makes all interest sums of
this kind straightforward When the rate of interest is other
than 6 per cent the method can sometimes be adapted
For instance to find the interest due on $284 for 54 days at
41 per cent

·92 Figuring Interest for 60 days at 6 per cent = 284 Interest for 60 days at 44 per cent = 2 13 (2 84 less a quarter) Interest for 6 days at 4 per cent = 21 Interest due for 54 days is therefore $2 13 less 21 cents - that 1s $1 92 COMPOUND INTEREST If the interest due on a loan is not paid, but allowed to accumulate, the sum invested will be larger at the end of each interest period When the interest and the original sum are 'compounded in this way the amount due at the end of each interest period rises - for interest is not only paid on the original sum, but on the interest already earned as well Interest calculated in this way is called compound interest Compound interest, like simple interest, is calculated by multiplying the principal by the period by the interest rate In the case of compound interest, however, the operation must be repeated for each interest period If $3000 is lent for a year at 4 per cent simple interest, payable quarterly the interest due at the end of each period will be $30, and the total interest earned over the year $120 Over two years it will earn twice that - $240 Now let us see what happens if the loan is for $3000, but at 4 per cent compound interest computed quarterly, and what is earned ov the same period of two years ·Percentages Discounts and Interest 93 Principal 3000 00 Interest for 1st quarter at 1 per cent 30 00 Principal at the beginning of 2nd quarter Interest for the period of 2nd quarter 3030 00 30 30 Principal at the beginning of 3rd quarter Interest for the period of 3rd quarter 3060 30 30 61 Principal at the beginning of 4th quarter Interest for the period of 4th quarter 3090 91 30 90 Principal at the beginning of 5th quarter Interest for the period of 5th quarter 3121 81 31 21 Principal at the beginning of 6th quarter Interest for the period of 6th quarter 3153 02 31 53 Principal at the beginning of 7th quarter Interest for the period of 7th quarter 3184 55 31 84 Principal at the beginning of 8th quarter Interest at the end of 8th quarter 3216 39 32 16 TOTAL principal at the end of two years 3248 55 The total interest earned is $3248 55 — $3000, that is $248 55 The difference between simple and compound interest in this case is therefore $8 55 There is no straightforward method of calculating compound interest mentally, alas although the shortcuts I have already described for working with percentages will often make the calculation quicker ·DECIMALS Decimals, as I explained in Chapter One, are one way of expressing parts of whole numbers To the left of the decimal point the value of each place is ten times greater than the value of the place to its right To the right of the decimal point each place is one-tenth of the value of the place to its left ten millionths 7 6 5 4 3 2 1 163 would be read as 16 and 3 tenths 16 38 would be read as 16 and 38 hundredths 16 384 would be read as 16 and 384 thousandths 16 3846 would be read as 16 and 3846 ten thousandths 16 38468 would be read as 16 and 38468 hundred thousandths 16 384631 would be read as 16 and 384681 millionths To shuft the decimal point one place to the left is to divide by 10 94

·Decimals 95
To shift it one place to the right is to multiply by 10 This is at
once very simple and a potential source of errors - a confusion
of one place about where the decimal point should go
will
give
you an answer too small or too great by a factor of 10 The rules
though are simple
In adding and subtracting see that the decimal points stand
directly below one another The decimal point in the answer
will come in the same place
393
9 0633
1 5161
303
468
99 7094
In subtracting you may have to add zeros to make up the number
being subtracted from to the same number of places as the
number being subtracted - or vice versa
936 from 41 0853421 becomes
41 0853421
9 3600000
31 7253421
or 9 364857 from 12 62 becomes
12 620000
9 364857
3 255143

·96 Figuring In multiplying carry out the calculation first and then find the place to insert the decimal point by counting as many places in from the right in the product as there are places in the multiplier and multiplicand together 9 4562 X70956 67 09741272 There are four decimal places in the multiplier and in the multiplıcand and eight therefore in the product When muluplying very small decimal fractions the zeros after the decimal point can be ignored until the final process of inserting the point in the product 000056 X 0923 First of all multiply 56 by 923 to obtain the product 51688 In the multiplicand there are six digits following the decimal point, and in the multiplier fou- digits Therefore the product is 0000051688 The same principle applies when multiplying more than two numbers together to multiply 3 15 by 315 by 31 5 first find the product of 315 x 315 x 315, which is 31255875 and count off six decimal places from the right - the total number of places in the three numbers being multiplied together In division too the same technique is used as with whole numbers The placing of the decimal point is however easier if the divisor or dividend is a whole number This can be achieved ·Decimals 97 by shufung the decimal point to the right Provided the point is moved the same number of places in both numbers the result of the division will be the same For example to divide 7 by 196 First muluply both the dividend and the divisor by 10 As both the numbers are multiplied by the same number, the quotient remains unaltered 7 becomes 7 and 196 becomes 1 96 Now carry out the division 190)700(3 57 5 88 I 120 980 1400 1372 28 And so on More genrrally the best rule is to move the dec mal point to the right in the divisor as many places as are necessary to make it a whole number Move the decimal point the same number of places to the right in the dividend Insert the decimal point in the quotient as soon as any figure is used from the decimal part of the dividend in working out the problem Some fractions cannot be resse decimals with a finite number of places, among these are fractions like 1/3 ( 3333333 etc) and 1/7 ( 142857142857142857 etc) which have single figures or groups of figures recurring infinitely But it is not only in cases like this that some decimal places are ignored When starting a calculation you should decide what degree of accuracy is required It is not necessary to continue any operation beyond ·a certain number of 98 Figurmg the third or fourth decimal place in most practical calculations To determine that a number is correct decimal places you must look at the rejected figures If the result 13 to be correct to four places you must know what number stands in the fifth place If it is 5 or greater add one to the preceding digit For example in the multiplication 13 064973 x 25 035 327 081599055 the answer runs to nunc places of decimals This, correct to one place of decimals is 327 1 for 8 is greater than 5 and 1 is added to the preceding figure Correct to two places the answer 327 08, and to three places 327 082 Correct to four places it is 327 0816 As the easiest mistake to make with calculations involving decimals is putting the decimal point in the wrong place it as always as well to make sure that you have a rough idea of the magnitude of the expected answer In an earlier example we multiplied 3 15 by 315 by 31 5 This is very roughly, 3 x 1/3 X 30, 1/3 of 3 is 1 and the ansaer should have two significant figures in the region of 30 followed by decimal places Having made thus rough check you can have confidence in the answer 31 255875 when you arrive at it ·12 FRACTIONS In the definitions at the beginning of this book I described fractions You will remember that a fraction has a numerator - which tells how many parts you have - and a denominator - which tells what proportion of a whole those parts are Thus in the fraction 3/4 the 3 tells how many parts there are, and thc 4 that cach is a quarter of a whole If the numerator and denominator are both multiplied by the same figure the resulting fraction will have the same value as the original one :-3/4 or 6/8 or 9/12 are all the same proportion of a whole It is usually casiest to deal with fractions reduced to their lowest terms, however, and if numerator and denominator have a common factor both are divided by it to achieve thus reduction 9/27 for instance is divided by the largest common factor of 9 and 27, which is 9, to arrive at the lowest terms of the fraction, 1/3 Common fractions fractions with a numerator and a denominator are either proper fractions those which have a larger denominator than numcrator, improper fractions - those in which the numcrator is greater than or the same as the denomunator or mured numbers - which consist of a proper fraction and a whole number Any improper fraction can also be written as a mused number and any mured aumber as an improper fraction, 21 for instance as 11/4 ·100 Figuring Addig fractions To add fractions we must find a common denominator That is to say we must add quarters to quarters and eighths to eighths, not quarters to cighths The simplest way to find a common denominator is to multiply the denomina tors together We Lnow that a fraction is not changed when its numerator and denominator art divided or multplied by thsame figure, so both numerator and denominator can, for each item in the addition be multiplied by the product of the other denominators For example 1 1 2 1 + المه The common denominator in this problem is the product of all the denominators in the original fractions That is 2 X 3 X 5 X 4 = 120 Thus the problem is represented as 60 120 40 120 t 48 120 + 30 120 60 + 40 +- 48+ 30 120 178 89 29 120 60 60 You will remember from Chapter 7 the procedures for find ing the least common multiple of a group of numbers The most convenient common denominator will be the least common multiple of the denominators of the fractions being added or subtracted in the example above for instance we could have uscd 60 rather than 120 as the common denominator If we add 1 3 4 + 1 3 + + 7 12 6 ·Fractions 101 the common denominator is 12, as it is the biggest denominator and each of the other denominators are exactly divisible into it We change the other denominators to 12 and the problem becomes 9 12 + 4 12 2 12 7 12 + 9+4+2+7 12 22 11 S == 12 6 When subtracting fractions you go through the same process of finding a common denominator and multiplying the numerators For instance to subtract 9/11 from 1/2 2 x 11 = 22 which is your common denominator The problem then becomes 7 18 ZZ 11 22 22 And addition and subtraction can of course be combined 1 3 + 3 5 2 3 4 Here the common denominator is 2 x 5 6 3 x 4 = 120 There fore the problem becomes 60 120 72 120 40 120 + 120 ED72 +0 - 99 120 82 41 120 When adding and subtracting mused numbers at a loss ·Ic2 Figars of ros dashuke u bole Cumbers Erst and then the fract For 42 pe 63 - 21 +11 31 becomes 6-2-1- } 1 1: 12 - 8 ++ 8 4 2 8 32+ * 12+*+ + + 2+++1+ 8 8 = 12+ 8 12 4 1 - 13 When muluplying by fractions the numerator of the prod is obtained by multiplying the numerators of the two fractu and the denominator of the product by multiplying! denominators. The product will of course be smaller than 1 two original fractions - oft is $ The method is very simp For example to find the product of 5/8 x 6/9 5 x 6 30 8x972 5 12 To divide one fraction by another - to find for instance ho many eighths there are in a half - invert the fraction you al dividing by the divisor – and multiply the numerators an denominators together as ar multiplication for cramp! 3/7 dinded by 619 3 9 3 x 9 75 ² 7x6 27 42 9 14 In multplications and divisions the work can be simpliſed af, whenever possible you cancel - that is divide a numerator and a ·Fractions 103 denominator by the same number For example 3/4 x 8/9 becomes 1/1 x 2/3, or 2/3 when the 4 and the 8 and the 3 and the 9 have been cancelled Cancelling can be done in more than one step a numerator may have different factors in common with different denominators and vice versa It can of course only be done when all the signs in the calculation are multiplication signs

·13
THE CALENDAR
In this chapter I will show you how to determine what day of the
week any date fell on or all fall on Any date that is, since
15 October 1582 when our present calendar the Gregorian
calendar Mas instituted
In order to do this mentally you will need to commit four
tables to memory The first is so simple it hardly justifies the
name Here it is
TABLE I
7 14 21 28
The second is more complicated
TABLE II
January corresponds to 0
February corresponds to 3
March corresponds to 3
corresponds to 6
May
corresponds to 1
June Corresponds to 4
104

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·106 Figuring At the end of your calculation you will have a figure which will correspond to a day of the rock in the following way 1 TABLE TV Sunday 0 Monday Tuesday 2 Yednesday 3 Thursday 4 Friday 5 Saturday 6 Toarnvc at this final figure you carry out the following operations i Reduce the date of the month to a figure less than 7 by subtracting the appropriate mulupie of 7 from Table I 2 To the figure thus obtained add the figure standing opposite the relevant month in Table II Again cast out the sevens 3 If the year falls after 1900 and is a leap year add the figure obtained after operation 2 to the figure standing opposnic the year in Table III, unless the month is January of February, in which case add one less than the figure shown If the scar is not a leap year add to the figurc obtained after operation 2 the figure standing opposite the previous leap year, plus the difference between that leap year and the year in question (If, for instance the jaar in question is 1942 add to 1 - which is the number standing opposite 1940- the difference between 1940 and 1942, to arrive at thc figure to add to the total obtained after operation 2) 4 Having arrived at the total of date month, and year higures cast out the sevens again (as in operation 1) to arrive at the figure which will correspond to the required day of the neck in Table IV ·The Calendar 107 The operations described here will give you the day of the week for any date in the 20th century (Remember by the way that 1900 is not a leap year, so there is no need to reduce by one unit for January or February ) To find what day dates between 1582 and the beginning of the 20th century fell on you need to add a century figure to the figure obtained for the equivalent year in the 20th century For the 19th century this is 2, for the 18th century it is 4 and for the 17th century 6 Here are some cxamples What day of the week was 9 April 1932? First 9 is reduced by subtracting 7 The date figure is therefore 2 In Table II the figure corresponding to April is 6 This is added to 2 2 +6 = 8 Cast out the 7, and you are left with 1 From Table III you find that 1932 stands against 5 5+1=6 From Table IV you get the day of the week you are looking for The number 6 signifies Saturday, 9 April 1932 was a Saturday What day of the week was 23 January 1940? Reduce 23 by casting out the sevens 23 - 21 = 2 Add the number for the month of January 2 +0=2 Add the number for the year 2+13 but in this case it is leap year and the month is January so subtract 1 3-1=2 whuch, by reference to Table IV, we find stands for Tuesday 23 January 1940 was a Tuesday ·108 Figuring What day of the week was 19 October 19427 Reduce 19 by casting out the sevens 19 - - 14 = 5 Add the number for the month of October 5+0=5 The leap year before 1942 - 1940 This has the number 1, which you add to the difference between 1940 and 1942 1 +2 = 3 This is added to the result of the previous operation 3+5=8 We cast out the sevens to get the answer 1, which tells us that 19 October 1942 was a Monday Table III can be extended into the future to give days of the week for any year in this century We will now look at some examples going further back - to work these out we still need the century figures as well as date, month and year numbers What day of the week was 9 August 18327 The date figure (9 - 7) 2 The month figure 2 The figure corresponding to the same year in the present century 5 The century figure 2 Total Casting out the sevens gives 4 9 August 1832 was a Thursday What day of the week was 2 October 1869? Date figure 2 Month figure 0 ·The Calendar 109 Year figure (the figure for the nearest leap year, '68, plus the difference bet ncen '68 and '69) Century figure 2 2 Total 2 October 1869 was a Saturday 6 What day of the week was 29 July 17447 Date figure (29 - 28) Month figure for July Figure of '44 Figure for 18th century 1 6 6 4 Total 17 Casting out the sevens leaves 3, 29 July 1844 was a Wednesday What day of the week was 26 December 16137 Date figure (26 - 21) Month figure for December Year figure for 13 (1 + 1) Figure for 17th century 5 5 2 6 Total 18 Casting out the sevens llaves 4, 26 December 1613 was a Thursday Once you have the tables by heart the calculations can be done mentally at very great speed EXACT AGES If you know a persons date of birth it should be very easy to

·110 Figuring
find out how old he is in ycars, months and days And sometimes
it is, for the subtraction can be done like this
What is the exact age of a person on 12 May 1973 if he was
born on 8 February 1936?
1973 5 12
1936 2 8
37 3
4
However, when there are more days in the date being subtracted
than in that being subtracted from, days have to be borroned
A month is taken from the bottom line and added to the top lire
For example
What is the cract age of a person on 12 May 1973 if he was
born on 17 March 19427
1973 5 12
19+2 3 17
17 is greater than 12 so days must be added to the top linc The
third month March, has 31 days, so 17 is subtracted from
31 + 12 to give the number of days - 26 Before subtracting
the months 1 must be added to the bottom line
The answer is 31 years 1 month and 26 days Or again
What is the exnct age of a person on 24 December 1972 of
he was born on 30 August 19397
1972 12 24
1939 8 30
33 3 25

·The Calendar 115
(31 is added to 24 in the top line in the days column, and 1
therefore to the months in the bottom line)
When the number of months in the top line is smaller than the
number of months in the bottom line add 12 to the top line, and
I to the years in the bottom line For cxample
What is the cruci age of a person on 19 August 1973 af he
was born on 16 October 19427
1973 8 19
1942 10 16
30 10 3
(12 is added to the months column in the top line, and I to the
years in the bottom linc) Or again
What is the cxact age of 1 person on 23 September 1973 1f
he was born on 29 November 19477
1973 9 23
1947 11 29
25 9 24
(30 is added to the days in the top line, and I to the months in
the bottom linc, 12 is added to the months in the top line and
1 to the years in the bottom line)

·SOME SPECIAL NUMBERS
In Chapter Twa, ukien ! *Maiting about the digies, I described
some of their special characteristics - the palındromes created
by the multplication of 1, 11, 111 eic by themselves for instance,
or the pattern thrown up when one adds together the digits of
products of a series of numbers nulupued by 4
Here I will say something more about such characteristics
The personalities of numbers and of families of numbers - arc
such that there are still mysteries to be probed
In writing of the number 6 in Chapter Two I said it was one
of the group of perfect numbers Perfect numbers are those
that are equal to the sum of all their divisors that are less than
themselves this is of curse not common 1+2+3=6,
but 1 + 3 is less than 9, and 1+2 +3 +4 +6 is greater
than 12 In fact the second perfect number is 28
1+2+4+7+ 14 = 28
and for the third you must go on to 496
1 +2 +4 + 8 + 16 +31 +62 + 124 +248 - 496
The fourth perfect number is 8128
112

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·114 Figuring made by the marbles as we built up the triangle numbers Here for instance is T4 Turn du triangle through 45 degrees and add another to it to form a rectangle 000 90000 00000 One side of the rectangle has the same number of marbles as the sides of thu triangk The other has that number plus 1 The numbe of marbles in the ectangle can be found by multiplying the number of marbles on the short side by the number on the long side - in the example 4 x 5 But the rectangle was made up from two identical triangles and the number of marbles in each can be obtained by dividing the total number of marbles in the rectangle by 2 This principle can be seen to work for all triangle numbers The T number shes the number of rows th rectangle will have one side with the same numbe of marbles as the T number the adjacent side will have one more The product of the T number and the T number plus 1, when divided by 2 gives the value of the T number in question for instance to find the value of TIS 15 x 16 2 120 ·Some Special Numbers 115 And to find T20 20 x 21 – 210 Or 1236 236 x 237 -=27966 2 142857 THE REVOLVING NUMBER I described some of the odditics of 142857 when discussing the digit 7 in Chapter Two Its habit of repeating itself in the total when the powers of 2 are multplied by 7 and added, after being set out in a staggered formation, for instance Here are some more Furst set out the digits in a circle Now muluply 142857 by the numbers from 1 to 6 142857 X1 142857 X2 142857 X3 142857 X4 142857 X5 142857 X6 142857 285714 428571 571428 714285 857142

·116 Figuring
You will sec that the number starts revolving - the same digits in
different combinations arrived at by starting from a different
point on the circle Multiply 142857 by 7 and things suddenly
change
142857
X7
999999
But there are still oddities in store, for instance of you mulaply
142857 by a really big number see what happens
1
142857
x 32284662474
4612090027048218
No resemblance to 142857 at first sight perhaps, but divide the
product up into groups of 6, 6, and 4 and see what happens
048218
090027
4612
142857
The revolving number appears again! Sometimes 142857 hides
deeper than that For instance in this muluplication the product
seems immune from 142857 extraction
142857
X 45013648
6430514712336

·Some Special Numbers 117
But divide it into 6, 6, and I in the same way as you did last tim
and you arrive at the following sum
712336
430514
6
1142856
Treat this total in the same way
142856
1
142857
And jou
winkle 142857 out
You can find oddities just by looking at the digits themselves
For instance if you divide them into two groups 142 and 857
the second figure of the first group, muluplied by the thund figure
of the first group gives the first figure of the second group
4 x 2-8
The sum of the first two figures of the first group gives the
second figure of the second group
1+4= 5
And the sum of all the three figures of the first group gives the
thurd figure of the second group
1+4+2=7

·118 Figuring To show the nort property of 142857 you have to draw up a table of the products of the number when multiplied by 1, 2 3 4 5 and 6 Horizontally and vertically the digits all add up to 27 7 2 3 8 2 4 5 7 8 8 2 7 8 7 5 4 2 5 1 7 2 8 4 1 8 27 27 27 27 27 27 1 5 4 7 5 2 27 27 27 27 27 27 And if you can t remember the number there is always an casy way to find it for 1/7 erpressed as a decimal 1s 142857 142857 142857 142857 and so on to infinity THE NUMBER 526315789473684210 This deceptively straightforward number is also circular in the sense that when it is multiplied by any number from 2 to 200 the product can always be read off from a circle made up of the figures of the multiplicand If the muluplicr is 18 or less the break comes after the digit which is the same as the multiplicr, or, where the multiplier has two figures the same as the last figure of the multiplier When the cut has been made and the two parts joined add a zero to get the final answer As all the digits in the number except 9 and 0 occur twice you must know which of the pair to make the break after The rule is this look at the figures following the two you are choosing between If the number you are muluplying by is between 2 and ·Some Special Numbers 119 9 make the break before the lower figure, if it lies between 11 and 18 make the break before the higher ore For example To multiply 526315789473684210 by 6 The break must come between 6 and 3 or 6 and 8 The multiplier lies between 2 and 9 so we make the break between 6 and 3 We write out the answer by taking the figures that follow 6 315789473684210, carrying on from the beginning of the number and joining on 526, and adding 0 to get the answer 3157894736842105260 Or to multiply 526315789473684210 by 14 The break can be made between 4 and 7 or 4 and 2 14 lies between 11 and 18 so we choose the higher number, and the answer can be read straight off 7368421052631578940 To multiply by numbers between 19 and 200 is a little more complicated If they are the numbers we have already dealt with multiplied by 10 - that is 20 30, 40 and so on up to 180 - there is no problem for we can carry out the procedure set out above and add 0 to the product Similarly to multiply by 200 carry on the operation as if multiplying by 2 and add two zeros to the answer 19, 38, 57, 76 and 95 give products in which all the digits are nines except the last two and (in all cases except 19) the first 526315789473684210 x 19 9999999999999999990

·120 Figuring
526315789473684210
X 38
19999999999999999980
526315789473684210
x 57
29999999999999999970
526315789473684210
X 76
39999999999999999960
526315789473684210
x 97
+9999999999999999950
These five numbers are special cases The rule for numbers
betucen 21 and 29 is this add 1 to the second digit of the
mulusplies Multiply the special number making the break
before the lower of the two possible figures When you rcach the
last digit reduce it by 1 Insert 1 at the beginning and 0 at the
end of the number to arrive at the final product For example
526315789473684210 x 27
Increase the second digit of the multiplier by 1 7+1 = 8
Now makc a cut between 8 and 4 in our special number and
write down the figures
421052631578947368
Reduce the last digit by one you have
421052631578947367
Annex a zero to thus number and attach the 1 at the beginning
to get the product,

·Annex a zero to this number and attach the number
Some Special Numbes 521
14210526315789473670
To muluply the special number by numbers between 31 and 37
use the same method as that set out above, but make the hreak
before the hugher of the two possible figures For example
526315789473684210 x 34
Increase the second digit of the multiplier by 1 to male 5 Male
the break between 5 and 7 to get the number 789473684210526313
Reduce the last digit by 1, put a 1 at the begintung and a 0 at
the end to arnie at the final product
17894736842105263140
For multipliers between 39 and 48 the method is the same as
for numbers between 21 and 29, but instead of I at the beginning
of the product write 2, instead of deducting 1 from the last figure
of the number deduct 2, and instead of adding 1 to the second
dignt of the mulnplier add 2 For example
526315789473684210 X 46
Add 2 to the second figure of the multipler 6+2-8 Make
a cut at the lower succeeding figure You get
421052631578947368
Reduce the last figure by two units You get
421052631578947366
beginning of the number to obtain the product

·122 Figuring 24210526315789 173660 When the product lies between 49 and 56 the method is the same as for the multiplicts princen 39 and 48 but the cut is made before the higher of he (0 possible figures When the multiplier is between 58 and 67 the method is as for those between 39 and 18 but where 2 was added in thosc calculations 3 is added The cut is made at the lower figure For multipliers between 68 and 75 the method is as for 58 10 67 but the break is made at the higher figure For multipliers between 77 and 85 insert of not 3, and cut as the lower figure and for multipliers becucen 86 and 94 do the same but cut at the higher onc For multipliers 96 to 104 insert 5 not 4 and cut at the lower figure For other numbers simular methods apply - you can find them by trial and crror, right up to 200 You will find though that 114 133, 152 171 and 190 guc products made up of a series of nines with two varying digits and a zero or zeros THE NUMBER 1089 This as I mentioned earlier (Chapter 8) has some peculiar traits rur instance look at the pattern that is formed when it is multplied by the numbers 1 to 9 1089 xls 1089 9801 = 1089 9 1099 x 2 =2178 8712 = 1099 y 8 1039 x 3 = 3267 7623 = 1089 <7 1089 X 4 = 4356 6534 = 1089 x 6 1089 x 5 = 545 Or try thus trick wnte 1089 on a piece of paper and put it in ·Some Special Numbers 123 your pocket Nox ask someone to think of a three-digit number the first and last digits of which differ by at least 2 I et us suppose he chose 517 Now ask him to reverse the digits and subtract the higher number from the lesser (715 – 517 = 198) Finally ask him to add this number to itself reversed (198_891 = 1089) No matter what number he starts with you will always have the final answer - 1089 - !n your pocket! THE NUMBER 12345679 This number too appeared earlier (Chapter 2) Now we will look at some of its other peculiarities Multiplying it by 3, and multiples of 3, grcs a curious set of patterns 12345679 y 3 = 37037037 12345679 x 6= 74074074 12345679 X 9 = 111111111 12345679 x 12 = 148148148 12345679 x 15 = 185185185 12345679 x 18 = 222222222 and so on Multiplying by 999999999 gives a number which is a mirtor image of itself 12345679 X 999999999 = 12345678987654321 Incidentally, this product is a perfect square 1111111111 12345678987654321

·124 Figuring
PRIME NUMBERS
A positive integer, a whole number, greater than one, that cannot
be expressed as a product of two positive integers, neither of
which is 1, is known as a prime number Any positive integer
greater than I that is not a prime number can be expressed as a
product of two or more prime numbers For example, 4 = 2 x 2,
6= 3 x 2,8 = 4 x 2 and so on
The first few prime numbers 2, 3, 5, 7, 11 etc, are easy to spot
Until 1952 the highest prime number known was 2117 – 1,
which written out in full is
170141183460469231731687303715884105727
Then five higher primes were found by computer the largest
being 22281 - 1 More recently a computer in Sweden has been
used to prove that 28871 – 1 is also a prime number
Two numbers that were thought to be primes on the other
hand have fately been shown to have factors They are
1757051 (1291 x 1361) and 222221 (619 x 359)
STRANGE ADDITION
1
2 6
3 9 15
4 12 20 28
5 15 25 35 45
6 18 30 42 54 66
7 21 35 49 63 77 91
8 24 40 56 72 88 104 120
9 27 45 63 81 99 117 135 153
In the table above the horizontal lines are arithmetical pro
gressions The defference between each number and the one to

·Some Special Numbers 125 its night is twice the figure that stands at the beginning of the row (Row 7 for cxample can be worked out by adding 14 first to 7 and then to cach successive total) But how would you find the sum of all the numbers in any row There is no need to add them - it is the same as the cube of the number which stands at the beginning of the row For instance the total of the numbers in tow 6 is om 6 X 6= 216 AND SOME STRANGE FRACTIONS Here are two fractions, with one as numerator and with two different uneven multiples of 9 as denominator 1 oxand 1 9 x 17 9 x 3 27 153 Now let us take the first fraction and reduce it to decimals, by dividing the numerator by the denominator 1/27= 037037037037 to infinity and take just two significant figures of the first three places of decimals 37 Now watch the pattern this number forms when mulupled by 3 and its muluples 37 x 3 =111 37 x 6 = 222 37 X 9 =333 37 x 12 = 444 37 x 15 = 555 37 X 18 666 37 x 21 = 777 37 x 24 = 888 17 po

·126 Figuring
Now take the second fraction and reduce st to decimals
1/153 = 0065359477124183
Tale all the significant figures, multiply by 17, and multiples of 17
and watch the pattern formed
65359477124183 x 17 1111111111111111
65399477127183 x 34 =2222222222222222
65359477124183 x 51 = 3333333333333333
653,9477124183 y 68
65359477121183 x 85 55555555
65359477124183 x 102 - 66666666
65359477124183 y 119 = 77777
65359477124183 x 136 = 8888888888888888
65359477124183 x 153 = 9999999999999999
REVERS ALS
9+9= 18
24 + 3 = 27
47 +2 = 49
497 +2 = 499
81 =
9x9
72 = 24 x 3
94 - 47 x 2
994 497 x 2
CASTING OUT THE NINES
In other chapters I have described ways of checking answers by
Casting out the nines Casting out the nincs - by repeatedly
subtrs aing 9 unul a remainder of Jess than 9 is lefi or, which
amounts to the same thing dividing by 9 ard noting tbc

·600045A '
BRE 2013 2012 'OS IN
(18612R 1RbJ9) wak
(1861 261 1962) je se bave HE
Some Special Numbers 127
r mainder - can be du cin an oddly simple way 1 he remainder
when a number has been duided by 9 is the same as the sum of
thr digits (or, when that sun gives a sumbue with no digits the
sum of those digits) As the remainder - not the numh r of nines -
is what you are after you can arrive at it directly 11 si are ino
exampies
Cast the mines from 67 and find the remainder
67
-9
9)5717
63
6 + 7 = 13
3+1=
4
40
-9
31
22
-9
13

·128 Figuring 9)4444 36 4+4= 8 35 8 26 17 8 PI Pr is the number which expresses the ratio of the circumference of a circle to its diameter While it has been proved (by Lindemann, in 1882) that it cannot be a repeating decimal (in mathematical terms that it is a transcendental number and not the root of a finite algebraic equation with integral coefficients), It has been calculated to a degree of accuracy that would once have seemed impossible "The story of the accuracy to which the value of Pi is known is an interesting one In the Bible the value of P1 given is 3 Archimedes declared the value of Pi to be less than 3 1/7 but greater than 3 10/71 The value generally used today 3 1416, was known at the time of Ptolemy (AD 150) In the 16th century r Vieta calculated Pi to ten places of decimals At the end of the 16th century, Ludolph van Ceulen calculated thirty-five decimal places in his will he requested that these thirty five numbers be engraved on his tombstone This was done, and this number is still known in Germany as the 'Ludolphian number, in lus honour ·Some Spenal Number: 129 Now computers have come to the aid of the mathematicians In 1949 a group using the rather primitive computer Entac took 70 hours to calculate Ps to 2037 decimal places, while recently Daniel Shanks and John Wrench Jos have published Pr 10 100000 places, arrived at in 8 hours and 43 minutes using an IBV 7090 system They used the following formula 3 1 24 tan-- + 8 can & 57 + 4 tan-? 238 I took them 3 hours and 7 minutes to compute 8 tan- 1/57 2 hours and 20 minutes to compute 4 tan-1/238 2 hours and 34 minutes to compute 24 tan-' 118, and 42 minutes for conversion (binary to decimal) Mnemonics are often used to help remember the digits of Ps In the sentence May I have a large container of coffee the value of Pa to seven places of decimals is contaired The number of lerrers in each word corresponds to the successive integers in the decimal expansion of P1 Sir James Jeans came out with the following sentence in which the value of P. is contained to fourteen decimal places "How I want a drink, alcoholic of course after the heavy chapters mzolving quantum mechuris 3 14 15 92 65 35 89 79 1 igest of Adam C Orn of Chicago published in the Chicago of 20 January 1906, the following Po to thirty decimal places Now I - esen In Rhymes Immortal S;

·130 Figuring
Who in his wonderous lore
Passed on before,
Left men hus guidance
How to circles mensurate
31415926535897543238452643383279
It is also said that this French poem contains P1 to thurty
places of decimals
Que j'aime a faire apprendre
Un nombre utile aux sages
Immortal Archimede Artiste Engenieur
Qui de ton jugement peut priser la valeur
Pour moi ton probleme
A les pareils avantages
However, there are two mustakes in the last line Where A les
represents the digits 1 and 3 the correct digits arc 3 and 2

·15 TRICKS and PUZZLES Creative mathematics is divided from play only very vaguely It is said that Leibnitz spent a considerable amount of time studying solitaire, and Einstein's bookshelves were stacked with books on mathematical games, so this section of tricks and puzzles may lead you to think more like a mathematician than some of the methods for solving standard problems in earlier chapters Much of science is after all finding mathematical solutions or descriptions for puzzles posed by nature 9992 Who Sits by Who Giss 28 It was a boarding school In the hostel dining-room the tables seated ten and on the first day of school, ten boys stood around one table and argued about who should sit next to whom One of the boys suggested that they should sit according to the alphabetical order of their names Another proposed that the seating should be planned according to their heights And yet another suggested that they should have a raffle and pick out the names against the numbers of the seats The argument continued Then the head boy of the class came out with a splendid proposal 'Let us all sit today according to the position we are standing around the table Tomorrow we will change the position The day after tomorrow ne again change the position This Way each day he will sit in a different order. And when we are back in the same position as we are sitting in today we shall celebrate It with a special dinner, and I shall stand the expenses This proposition seemed very reasonable to all of them and they agreed But the day never came when the head boy had to throw the 131 ·132 Figurmg special dinner party, even when they had all graduated and were ready to leave schooll There are too many different ways 10 boys could sit around a table The possible number of combinations 15 giten by the calculation 1 X2 X 3 X 4 5 X 6 X 7 X 8 X 9 X 10 = 3628800 If instcad of 10 there were 13 boys the possible combinations would have been 1x 2 X 3 X 4 X 5 X 6 X 7 X 8 X 9 X 10 x 11 x 12 x 13 = 6227020300 If there were only two boys the combination could have been just 1 x 2 = 2 If there were 3 boys 1x2x1=6 If there were 4 bays 1 x 2 x 3 x 4 = 24 If there were 5 boys 1 x 2 X 3 X 4 X 5 = 120 And if there were 15 boys the maximum number of combinations would be 1 X 2 X 3 X 4 X 5 X 6 X 7 X 8 X 9 X 10 * 11 x 12 x 13 x 14 x 15 = 1307674365000 If you consider 25 boys the number of combinations would be 1x2x3 x 4 x 5 x 6x7x8 x 9 x 10 x 11 x 12 x 13 x 14 x 15 x 16 x 17 x 18 x 19 x 20 x 21 x 22 x 23 x 24 Y 25 = 15511210013330985984000000 It would be cternity before a party of 25 people could exhaust all the combinations The Boy who could only Add Shambhu came to the city from a remote village Though he had never attended school he had been taught by his father how to add and divide by two This was all the anthmetic he had yet ·Tricks and Puzzles 133 he told a friend, he could rapidly multiply numbers with two digits The friend was surprised 'Very well,' he said, 'what is 43 times 877 Shambhu came up with the ansner, 3741, in a few minutes This is how he did it He took a piece of paper and wrote the number 43 on the left and 87 on the right He divided 43 by 2 and divided the answer by 2 again He continued dividing by 2 until he reached 1, ignoring the remainders at each stage except the last Then he doubled the 87 successively until he had an equal number of figures in ach column Having written these two columns of numbers he cancelled every number in the right-hand column that was opposite an even number in the left-hand column Then he added up the numbers remaining in the right-hand column to get the answer This is how his paper looked 87 174 43 21 10 5 2 87 174 696 2784 696 1392 2784 3741 1 Dividing the Elephants The Maharaja of an Indian State invited three artists to perform at his court He was much impressed, and honoured them with On the day of their departure he presented the three of them With seventeen elephants expressing a wish that the oldest of the gold and gifts ·134 Figurmg artists should get one half of the number of clephants, the middle one one-third of the number, and the last one one ninth When the artists returned home, they were absolutely puzzled how could they divide the clephants according to the wishes of the Maharaja? One-half of the seventeen elephants meant that the oldest artist gets cight and a half elephants What good is an elephant cut in tsol They argued and the leading citizens of the town acre called to sctile the dispute They tried in vain to carry out the will of the Maharaja for they found themselves constantly held up by the impossibility of making any share a n hole number of elephants They were about to give up in despair when someone in the crowd suggested that the temple priest, a wise, very learned man, be called in to settle the affair The temple pnest listened to the whole story in detail, got up from his seat and asked to be excused Then he went back to the temple and returned in a few minutes with the temple clephant Now," he remarked, 'the problem can be easily solved 1 sm temporarily gifting this temple clephant and pooling it in with the other elephants' There perc altogether eighteen elephants, with the temple elephant thrown in The priest addressed the oldest artist 'Here, you take away half the number of clephants he said and presented him with nint elephants Then he looked at the middle artist and remarked, 'You are entitled to one third of the number, and here are your sit elephants ' He gave away six elephants to him Now there Rere three elephants remaining including the templc elephant Then to the youngest artist he said 'Your share is one-ninth of the total number of elephants Therefore you get two elephants' He gave away iwo elephants to him The remaining clephant was the temple elephant 'I take back ·Tricks and Puzzles 135 the temporary gift I had made 'So saying he walked away with his clephant, back to the temple The will of the Maharaja was followed to the letter The Lily Pond In India water lihes grow extremely rapidly In one pond a lily grew so fast that cach day it covered a surface double that wluch It covered the day before At the end of the thirtieth day it enurely covered the pond, in which it grew But how long would it take two watcr lltes of the same size at the outset and at the same rate of growth to cover the same pond? Naturally, one is bound to think that if it takes one water lily 30 days to cover the pond, it should take two water lilies 15 days to do the same thing that is, half the time But this answer is wtong Since one plant doubles its surface every day, it will have half the surface left at the end of the 29th day Therefore with two water lilies at the outset, each will cover its half of the pond at the end of the 29th day, and not before Which Vacation ? It was vacation time Two men were travelling in a train The older man asked the young man seated opposite to him how old he was The young man replied, “The day before yesterday I was 19, and next year I will be 22' Can you guess during which Vacation and on what day of the month these men were It must have been the Christmas vacation and on 1 January, has birthday being on 31 December On 30 December he was stall 19 By 1 January he was 20 And on 31 December in the Same year he would be 21 and the next year on 31 December he travelling? 1 would be 22 ·136 Figuring As Old as the Year of Your Birth The grandson told his grandfather that in 1932 hc #75 as old as the last two digres of his birth year and half as old as the century in which he was born The grandfather replied that the same applied to him 100 That's quite inipossible said the grandson But the grandfather went ahead and prosed his statement Can you gliess how old the grandson and the grandfather sere? The grandson was born in the twenucih century and therefore the first two digits of his birth year are 19 The other two digits are half of 32, 16 The grandson was born in 1916 The grandfather was born in the suneteenth century The first two digits of his birth year arc 18 The rest of the digues doubled must be equal to 132 Therefore the number se kant is half of 132 And that number is 66 The grandfather was born in 1866 In 1932 he was 66 Broken Stones A store-kceper had a stone weighing 40 pounds which used to weigh goods in 40-pound lots But one day it fell and broke into four pieces He was about to dump them in the dustbın when a neighbouring store keeper stopped him and showed him five preces of stone he himself had which were the broken pueces of a stone that used to keigh 31 pounds He explained that with these five pieces he could weigh any whole number of pounds from 1 to 31, and advised him to check the picces before throwing them away The storc-keeper found to his surprise that with the four stones he could weigh any whole number of pounds from 1 to 40 What are the weights of the piece of stone used by the storekecper and his neighbour?

·Tricks and Puzzles 137 The store-keeper's 40-pound stone broke into pieces weighing 1, 3, 9 and 27 pounds His neighbour's 3l-pound stone broke into pieces werghing 1, 2, 4, 8 and 16 pounds Spreading Rumours It is amazing how fast rumours can spread Things witnessed by just one or two people became common knowledge within a couple of hours The speed with which rumours travel is extraordinary It is fantastic! But when we work the thing out mathematically, there seems to be nothing fantastic about it Let us examine a case and see how it works out Manju Guha comes from the city to visit her relatives living in a small town She arrives in their home in the morning at seten o'clock She has a very interesting piece of news from the city which she conveys to the family with whom she is staying There are 4 members in the family and it takes Manju Guha exactly 10 minutes to tell them the gossip Now can you guess how long it would take for this news to spread around all the 87,300 inhabitants of the town, if each person took just 10 munutes to tell it to another 4 people? 5 Manju Guha arrives at 7 am At 7 am the news is known to It takes her 10 minutes to convey the news to 4 By 7 10 am the matter is known to (1 + 4) Each of the four nastens to tell another 720 am the gossip is known to 5 + (4 x 4) By 7 30 am the gossip is Inown to 21 + (16 x 4) By 7 40 am the gossip spreads to 85 +(64 x 4) By 750 am to 341 +- (256 x 4) By 8 am to 1365+ (1024 x 4) non the rumour gathers momentum and So 21 85 341 3365 5461

·138 Figurmg
By 8 10 am is known to 5461 + (4096 x 4)
By 8 20 am the gossip spreads to 21845+ (16384 x 4)
21845
87381
The news that was known to only one person at 7 am is known
by everyone in town by 8 20 am 1 hour and 20 minutes later
Mathematically the whole calculation an be condensed into the
foloning addition
1+ 4+ (4 x 4) + (4 x 4 x 4) + (4 x 4 x 4 x 4) + (4 x 4 x
4 x 4 x 4+ (4 x 4 x 4 x 4 x 4 x 4)+(4 x 4 x 4 x 4 x
4 x 4 x 4) + (4 x 4 x 4 x 4 x 4 x 4 x 4 x 4)
that is 1 + 4 +47 +48 +44 +45 +4 +47 +41
which cquals
1
4
16
64
256
1024
4096
16384
65536
87381
The people of this particular tout not being gossip-mongering
types, conveyed the rumour to only 4 people cach But if cach
person had shared the rumour with more people it would have
spread with even greater speed!
How Many Tickets?
Imagine a small railway line with 30 railway stations Can you

·Can you combine four 4s with any other mathematical symbol
except numbers and produce all the whole numbers from 1
to 18? You may use the four symbols of addition, subtraction,
Tricks and Puzzles 139
guess how many different kinds of tickets the railway has to have
printed?
At each of the 30 stations the passengers can get tickets for
any of the other 29 stations Therefore the number of tickets is
30 x 29 =870
Proving the Impossible
Can you prove 1 =
27
Impossible you think? All you need is a very elementary
application of algebra
Ler
a=b
Then
axb-bxb
ab = b
Subtract a' from either side
ab - al=b' - a:
or
alb - a) = (b + a)(b − a)
Divide both sides by b - a
Then
a = b +a
Therefore
1=1+1
1 = 2
AU Fours

·140 Figurmg multiplication and division the square root symbol, and the decimal point You may, if you wish, put the fours side by sidc 44 - 4 2 10 4+4 4 4 4 4 x 4 4 +4 2 11 4 + 4 + 4 4 = 3 44+ 4 = 12 4 + 4 4 + 14 = 4 4 + 14 = 13 (***)+4-5 (4 x 4) + 4 4 4 + 4 + 4 + 4 = 14 4+4+4 V4 +4= 15 44 4 --- 4+4+4+4-16 (4 x 4) -(4+4) - 8 (4 x 4) + +-1 4x4+* = 18 4 +4+39 Imding the Counterfeit The gold merchant cxamined the nine gold coins hc had purchased at an auction that morning He had been told by the ·Tricks and Puzzles 141 auctioneer that one of the coins was counterfeit, and that the counterfert coin weighed a gram less than the genuine ones Hc was baffled as to how to pick it out His old clerk came to his assistance 'In just two weighings I can spot it,' he said And he did just that How did he do it? He split the nine coins in three groups of three coins each At the first weighing he took two groups and weighed them against each other If they balanced, it meant that the third group contained the counterfeit coin If they did not, whichever group weighed lighter contained the bad coin Now he Inew which group contained the counterfeit coin In the next weighing, he took two of the three coins from the group containing the counterfeit coin and weighed them against each other If they balanced the third coin was counterfeit Otherwise the lighter one in the balance was counterfeit Transformations Can you write 23 with only twos 45 with fours and 1000 with only nines? Yes! 22 + 2/2 44 + 4/4 999 + 9/9 The Millionaire and the Mierchant It was a beautiful afternoon, the Ist of June A millionaire walked into a pub to have a drink He was joined by a local merchant In the course of conversation, the merchant made a proposition He offered to pay the millionaire $1000 a day for the rest of the month starting from that very day, and in return he cxpected only pennies Just one cent on that first day, double that 2 cents the next day, 4 cents on the third day and so on doubling the amount crery day for the rest of the month ·142 Figuring The millionaire burst into loud laughter, thinking the merchant a fool, and agreed They shook hands and the deal was closed The merchant took out a bundle of $1000 and handed it over to the millionaire The millionaire gave the merchant a penny in returat It was agreed that they would meet every day in the same place and at the same time for the rest of the month, to carry out their bargain So the next day the millionaire cheerfully took the bundle of $1000 from the merchant and handed him over 2 cents And the next day again the merchant handed over the bundle of $1000 to the mocking millionaire and in return got his 4 cents in the first ten days the millionaire received $10000 from the merchant and in return had paid him the mere pittance of $10 23 But by the 17th day, the millionaire felt a little uncomfortable, when he had to pay 65 536 cents, that is $655 36 to the merchant in return for the $1000 re got from him for he saw that next day he must pay more than he received 131072 cents that is $1310 72 And when he looked to the future he got very worried indeed For day after day as the merchant handed over his $1000 he got back from the millionaire amounts many times more than that On the last day of the deal the 30th day, the millionaire had to Jiquidate his assets to the last cent in order to honour his agreement with the merchant Can you guess the amount of loss suffered by the millionaire? On the 1st day the millionaire pays I cent On the 2nd day he has paid 1+2 On the 3rd day he has paid 3 + 4 On the tih day he has pasd 7 + 8 On the 5th day he has paid 15 + 16 = 31 On the 6th day he has paid 31 +32 63 On the 7th day he has paid 63 +64 = 127 = 15 ·Tricks and Puzzles 143 The whole calculation can be represented as 1+2 +2+. 23 + 2 + 25 + 20 = 127 On the 8th day it is 1 + 2 +-22 +23 + 2 + 25 + 2 + 2? = 255 and so on On the 18th day he begins to pay more than he receives and on the last day the 30th of June the millionaire has to pay a sum of $5368709 12 The total amount paid by the millionaire to the merchant in the 30 days works out to (230 --I)cent 1073741823 cents = $10737418 23 whereas he had received from the merchant in the 30 days $30000 Therefore the loss suffered by the millionaire is equivalent to $10737418 23 $30000 = $10707418 23 Thousands by Eights Can you combine eight 8s with any other mathematical symbols except numbers to represent exactly 1000> You may use the plus minus division and multiplication signs In an earlier chapter of this book I gave one solution to this problem, but there are many more both simpler and more complex Here are some of them 8 +- 8 -(8 8 x 8 - 8) - 8 = 1000 8 8888 - - 888 8 =1000

·144 Figuring
- 1000
1995 )(8+889) = 1
868 x 8) + (8 + 8)) – 8–8–8=1000
(868+8) - 8+8+8) 8 -
8 = 1000
Telling and Not Telling
A girl, asked how old she was did not wish to glc out her age
direcily and yet did not want to be rude and rcfuse to give it
all Therefore she said, 'W1y age three years hence multiplied by
three, less three times my age three years ago will give you my
present age Can you guess her age?
Let us assume her present age is \ years Three years hence her
age will be X + 3 years Her age three years ago was X-3
years
361 + 3) - 3X - 3) = X
Therefore Y = 18 which is her age Three years hence she will
be 21 and three years back she was 15
(3 x 21) – (3 x 15) - 63 - 45 = 18
Guessing Birthdays
You can tell a friend the month and the date of his birth tory
easily
First of all ask him to keep in mind two numbers the numbes
of the month on shich he nas born and the number of the date

·Tricks and Puzzles 145 of the month (The months of course are numbered 1 to 12, from January to December ) Then you ask him to multiply the number of the month by 5 add to this 6 and multiply it by 4 and then add 9 Once again, ask him to multiply the number this time by 5, and add to it the number of the date on which he was born When he finishes she calculations ask him for the Gnal result and mentally subtract 165 from it. The remainder is the answer The last two digits of the number, give you the date of the month and the first digit or the frst two digits give the number of the month For example, if the final result is 1269 when you subtract 165 jou get 1104 From this number you know that he was born on 4 November The trick behind it is, of course the 165 The directions you give your friend are a disguised way of addıng 165 to the number of the month multiplied by one hundred When the number 165 is taken away from the tal, the number of the day, and one hundred times the number of the month are left! False Promises Some years back the following advertisement appeared in the popular press UNIQUE OPPORTUNITY! UNIQUE OPPORTUNITY! PRESSURE COOKER ONLY $1 DELAY MEANS DISAPPOINTMENT RUSH!!! AND WRITE FOR THE FREE APPLICATION FORM OFFER GOOD ONLY WHILE STOCKS LAST ·146 Figuring Housewives did indeed rush to write for free application forms Pressure cookers, in those days, were usually sold for $5 In return for her letter, what cach housewife received however was not an application form - but four coupons which she was told to sell to her friends at $1 each She was then to send the $4 thus collected, along with her own $1 to the company On receipt of this amount, the company sent her a pressure cooker And many housewives did get their pressure cookers for $1 - the other $4 came from the purses of their friends At first every housewife got her pressure cooker for only $1 and there seemed to be nothing fraudulent about the transaction The company kept its promise, and gave the housewives pressure cookers for $1 and yet was not losing a penny in the bargain They got their full price for the goods However, it was obvious that the whole thing was a kind of a swindle People, somewhere, must have been losing money Sooner than expected came the time when coupon holders found it impossible to dispose of the extra four coupons they had to sell if they were to claim their cooker The first group of house wives who got their coupons direct from the company, found no difficulty in finding buyers for them Each of these housewives was able to draw four new partners into the deal And each of these four new partners had had to dispose of their coupons to four other partners Let us calculate and see how fast the number of coupon holders increased 1 first round second round third round fourth round fifth round sixth round 4 20 100 500 2500 ·Tricks and Puzzles 147 set enth round eighth round minth round tenth round 12500 62500 312500 1562500 total 1953125 If everyone in a town of two million joined all would be covered by the tenth round But only one-fifth of them would get their pressure coolers The rest being left with valueless coupons Messing Numbers Here is a multiplication with more than half of the digits expressed by X's Can you spot the missing digits? XIX 3X2 X3X 3X2X X2X5 1X8X30 The last digit in the product is a zero Therefore the digit in the thurd line has got to be a zero The X at the end of the first line must be a number that gives a number ending with zero if multiplied by 2, and with 5 if multiplied by 3 as the number in the fifth line ends with a 5 Only 5 will do this The X in the second line must be 8 - that is the only number which, when mulupled by 15, gives a number ending with 20 ·148 Figuring in the fourth line It is clear that the first X in line 1 us a 4, because only 4 multiplied by 8 grues a number that begins with 3 in the fourth line Now there will be no difficulty in spotting the remairung digits 415 382 830 3320 1245 158530 Thirty Reporition Can you write 30, using any three identical digits excepe 557 Here are three solutions 6x6-6=30 33 + 3 = 30 33 -3 = 30 Driding the Cake Three mon sere travelling in a railway compartment One of them Kas a very wealthy businessman The other two had food with them and invited the businessman to join them in their meal One man had brought 5 chapathis and the other had 3 The three of them ate together and the food was shared equally On parting the businessman gave 8 gold coins to the two men as a token of his appreciation of their hospitality, and told them to sharc the coins in proportion to their contribution of food An argument started The man who had brought 5 chapaldus felt that he should get 6 coins and the other man should get 2 ·Tricks and Puzzles 149 coins But the man who had brought 3 chapathis felt that he should get 3 coins and the other man should get 5 coins As the argument continued a crowd gathered Then a wise old man, who was listening, inten ened and offered to settle the dispute He heard both sides of the case and declared that the man who had brought 5 chapathis should get 7 coins and the man who had brought 3 chapathis should get only one The people in the crowd protested and demanded an explanation His explanation was this the food was shared equally by three men Each chapathi was portioned into three equal parts Therefore the 8 chapathus together were portioned into 24 equal pieces of the 24 preces, each man ate 8 preces The man who had brought 5 chapathis contributed 15 pieces, and the man who had brought 3 chapathis contributed 9 pieces Out of the is preces, the man who had brought 5 chapathis himself ate 8 pieces and contributed to the guest 7 pieces And out of the 9 pieces, the man who had brought 3 chapathis, himself ate 8 pieces and contributed to the guest only 1 piece So the man with the 5 chapathis gets 7 coins and the man with 3 chapathis gets only 1 All the Digits Do you notice anything interesting in the following multiplicatron? 138 x 42 =5796 There are rune digits and all are different Can you think of other such combinations? Here is a group of nine

·150 Figurmg
186
12 X 483 =5796
x 297 = $346
7254
, 48 x 159 = 7632
27 x 198 = 5346
28 y 157 =4396
4 x 1738 = 6952
4 x 1963 = 7852
Another Impossible Proof
Can you prove 45 – 45 – 45?
Write down the figures from 9 to 1
987654321 add up the digirs
9+8+7+6+5+4+3+2+1 = 45
Subtract 123456789 Add up the digits
1+2+3+4+5+6+7+8+9=45
Result
864197532 Add up the digits
8+6+4+1+9+7+5+3+2 = 45
Therefore 45 - 45 = 45
Dwindling Gifts
Two fathers gave their sons some money One father gave his
son $300 and the other son gave his son $150 Then the two sons
counted up their money, they found that berseen them they had
only $300 How do you explain this
While there are two fathers and two sons, there are only three
people - grandfather father and son The grandfather gave his
son $300 Out of this $300 the father gave his son $150 Thu
between the father and the son - two sons - they have only $300

·Tricks and Puzzles 151 Making the Signs Right 3328 = 18 This looks absurd, but it can be made sense of by the insertion of mathematical signs, without altering any of the figures, or the position of the equals sign Here is the solution V332 - 8 = 18 Fre Digits Can you arrange the digits 1, 2, 3, 4 and 5 in such a way that, with the aid of simple mathematical signs, they will make 111, 222, 333 and 9997 Here is the solution 135 - 24 = III 214 + 5 + 3 = 222 345 - 12 333 (58 x 4 x 2) - 1 = 999 Mlakmg One Hundred Can you, with only the digits 1,2,3 4 5, 6, 7 and 8 and the addrtion and subtracuon signs make 100% 86 +2+4+5+7-1 - 3 = 100 Smallest Integer Can digits? you spot the smallest integer that can be written path ·by comparateten digues? 148 35 29670 All Nmes Using only five nines, can you write 10? There are several ways of doing this 9 9+ 99 99 = 10 99 9 * 10 (9+ $)' 99 10 One Hundred from Ten Can you write 100 by using all the ten digits Give four different solutions 3 9 70 +- 24 18 + 5 I 38 50 + 49 2 =100 4 12 87 +92 + 3 +of 100 19 100

·Tricks and Purcks 153
Mfagu and Thirty-seren
Weite 37 on a piece of paper and lecp it in your pocker Ask
someone to suggest any number formed of three identical figures
Ask him to add up these three figures and divide the oriminal
number by the answer
Whatever number he chose you will always has the answer -
37 - on the paper in your pocket
555 divided by 5 + 5 + 5 gases 37
999 divided by 9 + 9 + 9 gives 37
888 divided by 8 + 8 + 88110s 17
and so on.
Hare and Tortoise
Two cars do the journey between cities 200 miles apart Omn car
# S0 mulos an hour one way and 40 miles an hour on the melon
Journey The other car at 45 miles an tour on bo hju TcYX.
Which of the two cars covers the distance 17 less time?
The speed of the second car is the average between the sport
ofthe first and second journeys of pha fost car But that dari
cover the distance in the same time
The first car tales on its first journes 200 – Oost for at
journey
200 - 40 = Stours Italibu
puncys of 9 hours Tře sccord car, my bush plotme eines
40 - 48 hours and 53 migues
This sccond ar covers the distarce in a shopping
tud on its second
Threes
you take folt 3s and simple matome vald meandem apo
Bean that they will reis in it - 11 si are!!!

·33
37
52
3
ما
3
3
The Chess Bour]
Perhaps the most famous of the striping about people who are
caught out by the way repeated doubling qurily leads to Alge
totals is that of hing Purushottama and the learned Pandit
The king proud of his abilities at chess had challenged ard
beaten all comers until afte , long strumile the Pandit bes!
him
The apparently insigtuificant reward he asked was l grain of
nice for the first square of the chess beard 2 for the scord
4 for the third 8 for the fourth and so on
When the sack of rice was slow in coming the Ling alled for hes
ministers and asked why
The total number of grains - 18H6744073709351615 - #35
not only beyond thc means of his granaries bur beyond the means
of the granaries of all the world The formula fur artmın, ar this
number - 24 - I - turns up in another famous Icgend
The Tower of Brahma
This so the legend has it is a temple in Benaras They
when the great lord Brahma created the world he put up three
say
that

·Tricks and Puzzles 155 diamond sticks, mounted on a brass plate, beneath the dome that marks the centre of the world Upon one of the sticks he placed 64 gold discs The biggest at the bottom, ali decreasing in size with the smallest at the top The temple prests have the tasks of transposing the discs from one stick to the other, using the thud as an ard They work day and night, but must transpose only one disc at a time and must not put a bigger disc on top of a smaller one When their task is complete, the legend says, the world will disappear in a clap of thunder Now the total number of transpositions needed to shift the 64 discs is, again, 204 – 1or 18446744073709551615 If every transposition takes a second it will still take 500000 million years to get the job done 1729 This is known as Ramanujam's number When the great Indian mathematician was lying ill in hospital, Professor Hardy, who had been instrumental in bringing him to Cambridge, came to visit hım The taxı I came in, he said to Ramanujam, had a very boring number - 1729 Ramanujam s face lit up No, he said, that is not a boring number at all, it is the only number that is the sum of two cubes in two different ways 10* + 98 = 1729 and 123 + 18 = 1729 Dimensshing Remamders Can you name the smallest number, which when divided by 10 gives a remainder of 9, when divided by 9 gives 8 as the remainder, when divided by 8 gives 7 as the remainder and so on, down to a remainder of 1 when divided 27 If X is the number, the problem can be set out thus

·ultiple of 10
mltiple of 9 -
Inple of 8-1
muinple of 7-1
multiple of
* mul iple of 5 - 1
- mulupic of -1
I multiple of 3 - 1
multiple of
Therefore
multiple of 10
\ in multiple of 9
multiple of 8
1.1 multiple of 7
VH1m mulrisic of 6
\+1 * muluple of 5
multiple of 1
muluple of 3
mulupic of 2
Now he find the LCM of the ten numbers That gives
2' x 31 x 5 x 7 2520
Therefore the required numbce is
2520 - 2519

·CONCLUSION Mathematical discoveries like all others may come from the need to solve practical problems But a more important motive is the pure delight some individuals take in invention and discovery A mathematician tries to extend his powers He knows a trick which allows him to solve one sort of problem - but might there not be other sorts that can be solved in the same way? He tries to find out how his tricks work, to group problems in families to classify them When a new problem comes up he will often be able to see immediately what family it belongs to, and what methods are likely to solve it In this book I have tried to arouse this active state of mind to encourage mental adventurousness If my readers have come to share this feeling for the subject my objective has been attained 157

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