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.

10804.. 74

78

86436

75635

We proceed, saying, eight times four make thirty-two, and four we carry from the remainder make thirty-six; eight times nought but three; eight times eight are sixty-four; eight times nought but six; eight times one are eight. The next 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;

842786

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.

Addition.

61 74

36

97

Take the fifth sum; instead of saying two and six make eight, and seven make fifteen, and six make twenty-one, and four make twenty-five, and one make twenty-six, write six and carry two; you must learn to think without repeating, even in your own mind, the amount of any two figures, merely repeating the amount without naming the figures. I take the sum, beginning 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.

46

32

346

270

191

Take the second example; instead of saying one from nought I cannot, but one from ten and nine, carry one to nine makes ten, and so on, the whole process must arise as it were an impulse of the mind; at least you should never repeat more than these words, one from ten and nine, ten from seventeen and seven.

79

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.

796

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 five make eighty-six and six make ninety-two; nine times seven are sixty-three and nine are seventy-two, and nine make eighty-one, and seventy make one hundred and fifty-one, the left figure being counted tens, and added.

19

15124

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

8) 4965000

620625 product.

were taught in a former lesson that to multiply any sum by ten 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

more than the proper multiplier; if therefore we take from this product once the multiplicand, the remainder will be the product required.

276

27324

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 the next two figures thirty-six, the product of the six is therefore multiplied by six, whose product is the same amount as 4963 would be if multiplied by 36.

4963

366

29778

178668

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 Long Division; but six times six being thirty-six, if divided as before described, the quotient is the same as if divided by thirty-six by Long Division.

6) 973

6) 162.. 1

27.. 1

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, 9, 11, numbers whose product is 495. Now let us see how this is applied to the rule of which we are treating.

Divide 79695 by 495.

5) 495

9) 99

11

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.