Exemplification for the Black-board.

Short Division; that is, where the divisor does not exceed 12.

4. Divide 63543 by 4.

Divisor, 4)63543 Dividend.

Quotient, 15885" 3 undivided remainder.
Divisor,

4

Proof, 63543

Suggestive Questions.— How many fours in 6 of the fifth rank? The 2 that remain of the fifth rank make how many of the fourth rank? How many fours in 23 of the fourth rank, then? The 3 that remain of the fourth rank make how many of the third rank? How many fours in 35 of the third rank, then? How many of the second rank are the 3 that remain of the third rank? How many fours in 34 of the second rank, then? How many of the first rank are the 2 that remain of the second rank? How many fours in 23, then? Our quotient, then, appears to be 15885 and 3 remainder. Is the remainder carried to the right or to the left in division, as above? Which way are numbers carried in all other operations; that is, in addition, subtraction, and multiplication? Why, then, do we commence at the left in division, and at the right in all other operations?

Proof. In division, what terms are factors of the dividend? See p. 57, 5. What terms, then, multiplied together, will reproduce the dividend, if the work be correctly done? Is the remainder a part of the quotient? Is it also a part of the dividend? Why must it be added in when the dividend is reproduced by multiplying the divisor and quotient?

5. Divide 264852 by each of the digits severally from 2 to 9, also by 11 and 12, proving each problem by multiplication, as above.

6. Divide 65382432 by each digit separately from 2 to 9, also by 11 and 12, and prove by multiplication.

7. Divide 97862432 by each digit separately from 2 to 9, also by 11 and 12, and prove by multiplication.

Exemplification for the Black-board.

Long Division; that is, where the divisor exceeds 12.

8. Divide 64235 by 24. [Place the three methods together on the black-board.]

Dividend,

a. The Long Method.

1st partial product, 48000 2000

64235(24 Divisor.

600

Partial Quotients.

2d partial product, 14400

6

b. Contracted Method, by omitting unnecessary ciphers.

c. Abridged Method, by performing the Subtraction mentally.

Dividend, 64235(24 Divisor.

11 undivided remainder.

Solution. The contracted method differs only in one respect from the exercises already performed by the pupil in division, and that is, by writing down each partial product before subtracting it, instead of performing the subtraction mentally. In the abridged method, resort is again had to mental subtraction. Take notice, however, that it is frequently necessary to add more than one ten to the minuend and subtrahend. Thus, in dividing the first partial dividend, we have 6×4=24, which cannot be taken from 2; adding 30=32 leaves 8; 6×2=12, and adding 3 (namely 30 of the rank on the right)=15 from 16-1. In dividing the second partial dividend, we have 7X 4=28, which cannot be taken from 3; adding 30=33, leaves 5; then 7X2=14, adding 3 (30 of the rank on the right)= 17 from 18=1, and so on, adding always as many tens as may be necessary. The Long Method is merely given to show the reasons for the different steps in the Contracted and Abridged Methods.

After the class have studied the exemplifications in the book, they should be written on the black-board; the first in full with all the explanations; the second with the figures only; of the third, merely the divisor and dividend. The divisor is placed at the right, not only to save room, but to bring together the factors of the dividend for the proof. But the mind of the pupil should not be shackled by confining him to particular forms. It will frequently be found convenient in practice to be able to place the terms in division in various positions, as below. That he may not be cramped by forms, let him practise with those different positions in school.

Divisor,

or Divisor) Dividend (Quotient

Divisor,
Dividend(or Divisor.
Quotient.

Questions for the Contracted Method. - What is the upper

number on the left? [Point to the dividend.] upper number on the right? What is the first

What is the number under

the dividend? Of what numbers is it the product? To what rank does it belong by position? [Point, if necessary, to the corresponding number in the Long Method.] What is the next number below? Whence comes the 2 on the right? What is the next number below? [Always pointing.] It is 144 of what rank by position? [Here point again, if necessary, to Long Method.] What is the next number? Whence the 3? What is the rank of this number by position? What is the next number below? It is 168 of what rank by position? What is the next number? Whence the last 5? It is 155 of what rank by position? What is the next number? Of what rank, and why? What is the next number? An undivided remainder from what? Why is it placed in the quotient with. the divisor written under it? To show that it has not been Will the product of the divisor and the integers in the quotient be the exact dividend? What number is necessary to complete the dividend?

[Let the teacher now write out and explain on the black-board the Abridged Method, as given above, and then call on one or more of the class to perform similar operations, and give similar explanations on the board.]

It may be proper to observe here that, when the divisor consists of many figures, the pupil at first may not readily ascertain how many times it is contained in the partial dividend. To obviate this difficulty, all the figures in the divisor, except the first [or the first and second], may be neglected for the moment, taking care, however, to make proper allowance for them, especially for the second [or third] figure. Thus, supposing the question was how many times 356 is contained in 932, the 56 may be neglected, provided it be remembered that the 3 (the first figure) represents for the moment more than 31, consequently can only be contained twice, and not three times, in 932. Again, supposing the first two figures of the divisor to be 48 or 49, it should be recollected that the first figure is in fact nearer to 5 than 4. But, in spite of the utmost care of a learner, a wrong figure will occasionally appear in the quotient. For division must AT FIRST be merely a series of suppositions and trials. Happily a correction is easily made on the slate. If, then, the remainder proves to be equal to, or greater than the divisor, the quotient figure must be too small, since the remainder shows that the divisor is contained in the partial dividend a greater number of times. On the contrary,

if the product of the quotient figure and the divisor be greater than the partial dividend, it is equally evident that the quotient figure is too large. In either case the correct number must be substituted. Beginners may be materially assisted by forming a table of products of the divisor with each separate digit except the first, as in the following example, according to the contracted method.

But aids of this sort should be as seldom used as possible, and when used, should be soon discontinued.

Sometimes it happens, after the figure has been brought down from the dividend to the remainder, that the number is still too small to contain the divisor, as in the following example.

Dividend, 80520(264 Divisor.

132 3 Part of the quotient.

Here 264 was found to be contained 3 times in 805 with a remainder of 13. Bringing down the 2 from the dividend, the number 132 is found to be too small to contain the divisor. As this shows that there are no tens in the quotient, a cipher should be put in the place of tens. For, in fact, 264 goes no times in 132, and leaves the same number (132) as a remainder. Bringing down, therefore, the last figure of the dividend (the 0), we ask how many times 264 is contained in 1320, and finding it to be 5 times, the 5 is placed as usual in the quotient, making it 305.