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while division is used to separate or diminish-one rule will prove the other.

Take the second example in Multiplication; we find the product is 18935703, and that the multiplier was 249; now place these two, the one as the dividend, the other as the divisor, and the quotient will be the multiplicand 76047.

249) 18935703 (76047

1743..

1505
1494

1170
996

1743
1743

There is another method of proving this rule, but as there is no certainty of its correctness, it is useless to burden the pupil's mind with it.

To prove any sum in Division : Multiply the divisor and quotient together, taking in or adding the remainder, and the product will be the dividend.

Take the first example in Long Division, and multiply the quotient 10804, remainder 74, by the divisor 78. We proceed, saying, eight times four

10804.. 74 make thirty-two, and four we carry from 78 the remainder make thirty-six ; eight times nought but three ; eight times eight

86436

75635 are sixty-four ; eight times nought but six ; eight times one are eight. The next 842786 figure of the multiplier being 7, we say, seven times four are twenty-eight, and seven, the figure in the remainder, make thirty-five, which we write in full, the next figure of the multiplicand being a cipher ;

and say, seven times eight are fifty-six, which is also written in full; seven times one are 7. Thus our product, 842786, is the dividend of the sum we wished to prove.

Having now gone through the fundamental rules, and given directions how they are to be proved, it only remains to point out to the learner what is the best mode of abbreviating or shortening the work in them. In all the rules it is necessary that the learner should discontinue, as far as he safely can, the use of those words by which he was taught to recite his work.

97

Addition, Take the fifth sum; instead of saying two and six make eight, and seven make fifteen, and six 61 make twenty-one, and four make twenty-five, 74 and one make twenty-six, write six and carry

36 two; you must learn to think without repeating,

46 even in your own mind, the amount of any two

32 figures, merely repeating the amount without naming the figures. I take the sum, beginning

346 at the right, and say, eight, fifteen, twenty-one, twenty-five, twenty-six ; at the bottom again saying, five, nine, eighteen, twenty-one, twenty-eight, thirty-four.

Subtraction.
Take the second example; instead of saying

270 ore from nought I cannot, but one from ten and

191 nine, carry one to nine makes ten, and so on, the whole process must arise as it were an

79 impulse of the mind; at least you should never repeat more than these words, one from ten and nine, ten from seventeen and seven.

Multiplication. There are many sums in this rule that can be abbreviated by methods peculiar to themselves, as well as by omitting those terms that apply to all rules generally.

Sums whose multipliers are between twelve and twenty can be multiplied in one line, by adding the figures of the multiplicand in with the product of the next figure on its left. Thus, we say nine times six are fifty-four, nine times nine are eighty-one and

796 five make eighty-six and six make ninety-two; 19 nine times seven are sixty-three and nine are seventy-two, and nine make eighty-one, and

15124 seventy make one hundred and fifty-one, the left figure being counted tens, and added.

To multiply by twenty-five, write two ciphers on the right of your multiplicand and divide by four. To multiply by 125, add three ciphers and divide by eight.

Multiply 4965 by 125.—You were taught in a former lesson 8 ) 4965000 that to multiply any sum by ten

620625 product. you added one cipher, by a hundred two ciphers, and by a thousand three ciphers; which latter was the process adopted in this instance. I multiplied it first by a thousand, which made the product eight times more than I wanted for a hundred and twenty-five, which is only the one-eighth of a thousand. I therefore divided the product of the sum multiplied by one thousand, by eight; and the quotient is the product of 4965 multiplied by 125.

When the multiplier is 9, 99, 999, or any number of nines, the sum may be worked by adding as many ciphers to the multiplicand as there are nines in the multiplier, and from this product subtract the multiplicand.

Multiply 276 by 99.-By adding two ciphers 27600 we multiply by one hundred, which is one

276 more than the proper multiplier ;

if therefore we

27324 take from this product once the multiplicand, the remainder will be the product required.

When it happens that some portion of the multiplier is a composite number, and that another portion is a factor of the same number, the work may be abridged thus. Multiply 4963 by 366. In this sum the unit six is a factor of 4963

366 the next two figures thirty-six, the product of the six is therefore multiplied by six,

29778 whose product is the same amount as 178668 4963 would be if multiplied by 36.

1816458

Division. Long Division may be made more easy by taking two or more figures, that, when multiplied together, will produce the divisor, and then dividing them in succession each in the quotient of the other.

Divide 973 by 36.
This commonly would be worked by

6) 973 Long Division ; but six times six being

6) 162.. 1 thirty-six, if divided as before described, the quotient is the same as if divided by

27..1 thirty-six by Long Division.

The technical terms used to express the above figures are multiples and submultiples ; thus thirty-six is the multiple, the two sixes are the submultiples or factors, and simply means that if two sixes are multiplied together they will produce thirty-six ; but after the multiplication has taken place they change their names, and

are then called the product and its factors. If I ask what submultiples would produce the multiple thirty-six, the answer would be two sixes, being the factors, the product is thirty-six, I have said this much about multiples and submultiples for the purpose of enabling the pupil to find the submultiples of any sum he may wish to divide by.

Find the submultiples of 495. Thus we have the submultiples, 5,

5) 495 9, 11, numbers whose product is 495.

9) 99 Now let us see how this is applied to the

11 rule of which we are treating.

Divide 79695 by 495.
(1.) 5) 79695 (2.) 495) 79695 (161
9) 15939

495••
11) 1771

3019

2970
161

495
495

On the right the sum is done by Long Division, on the left by a continued division of the factors or submultiples of the divisor; but recollect that no sum can be so done in this rule whose divisor is not evenly divisible.

To discover more readily the submultiples of any sum, you will recollect that any even number is divisible by 2; that four will divide any sum, however large, whose last two figures are evenly divisible by it; and that 6 and 8 have somewhat similar powers; and that all numbers ending with 5 or 0 may be divided by 5. This leaves but 3, 7, 9, 11, and you will readily perceive what numbers are evenly divisible by these, as they arise in the progress

of

your work.

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