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quotient multiplied by the divisor produces the dividend, it is evident, that if the product of two numbers be divided by one of those numbers, the quotient must be the other num ber.
It appears that division is applied to two distinct purposes, though the operation is the same for both. The object of the first and fourth of the above examples is to divide the numbers into equal parts, and of the second and third to find how many
times one number is contained in another. At present, we shall confine oui attention to examples of the latter kind, viz. to find how many times one number is contained in another.
At 3 cents apiece, how many pears may be bought for 57 cents ?
It is evident, that as many pears may be bought, as there are 3 cents in 57 cents. But the solution of this question does not appear so easy as the last, on account of the greater number of times which the divisor is contained in the dividend. If we separate 57 into two parts it will appear more easy.
57 = 30 +-27. We know by the table of Pythagoras that 3 is contained in 30 ten times, and in 27 nine times, consequently it is contained in 57 nineteen times, and the answer is 19 pears. Horo
many barrels of cider, at 3 unllars a barrel, can be bought for 84 dollars ?
84 = 60 + 24 3 is contained in 6 twice, but in 6 tens it is contained ten times as often, or 20 times. 3 is contained in 24 eight times, consequently 3 is contained 28 times in 84. Ans. 28 barrels. How
many pence are there in 132 farthings ? As many times as 4 farthings are contained in 132 farhings, so many pence there are.
132 = 120 + 12 120 is 12 tens, 4 is contained in 12 three times, consequently it is contained 30 times in 12 tens. 4 is contained 3 times in 12 units, consequently in 132 it is contained 33 times. Ans. 33 pence.
How many barrels of flour, at 5 dollars a barrel, may be bought for 785 dollars.
Operation. 785 = 500 + 250 + 35 5 is contained in 5 once, and in 500 one hundred times. 250 is 25 tens. 5 is contained 5 times in 25, consequently 50 times in 250. 5 is contained 7 times in 35 units. In 785, 5 is contained 157 times. Ans. 157 barrels.
How many dollars are there in 7464 shillings ?
As many times as 6 shillings are contained in 7464 shillings, so many dollars there are.
Operation. 7464 = = 6000+ 1200 + 240 + 24 6 is contained 1000 times in 6000, 200 times in 1200, 40 times in 240, and 4 times in 24, making in all 1244 times.* Ans. 1244 dollars.
It is not always convenient to resolve the number nto parts in this manner at first, but we may do it as we perform the operation. In 126 days how many weeks ?
126 = 70 + 56 Instead of resolving it in this man ner, we will write it down as follows. Dividend 126 (7 Divisor
10 56 8 56
e observe that 7 cannot be contained 100 times in 126, I therefore call the two first figures on the left 12 tens or 120, rejecting the 6 for the present. 7 is contained more than once and not so much as twice in 12, consequently in 12 tens it is contained more than 10 and less than 20 times. I take 10 times 7 or 70 out of 126, and there remains 56, Then 7 is contained 8 times in 56, and 18 times in 126. Ans. 18 weeks.
* Let the pupil perform a large number of examples in this manner when he first commences; as he is obliged to separate the numbers into parts, he will at length come to the common method.
In 3756 pence how many four-pences ?
It is evident that this answer will be obtained by finding how many times 4 pence is contained in 3756 pence.
If we would solve this, as we did the first examples, it will stand thus :
3756 3600 + 120 + 36 But if we resolve it into parts, as we perform the operation, it will be done as follows: Dividend 3756 (4 divisor
900 = number that 4 is contained in 3600 156 30 do.
120 120 9 do.
Here I take the 37 hundreds alone, and see how many times 4 is contained in it, which I find 9 times, and since it is 37 hundreds, it must be contained 900 times. 900 times 4 is 3600, which I subtract from 3756, and there remains 156. It is now the question to find how many times 4 is contained in this. I take the 15 tens, rejecting the 6, and see how many times 4 is contained in it. It is contained 3 times, and since it is 15 tens, this must be 3 tens or 30 times. 30 times 4 is 120. This I subtract from 156, and there remains 36. 4 is contained in 36, 9 times ; hence it is contained in the whole 939 times. Ans. 939 four-pences.
If these partial numbers, viz. 3600, 120, and 36, are compared with the resolution of the number above, they will be found to be the same.
This operation may be abridged still more.
In this I say, 4 into 37 9 times, and set down the 9 in the quotient, without regarding whether it is hundreds, or tens, or units, but by the time I have done dividing, if I set the other figures by the side of it, it will be brought into its proper place. Then I say 9 times 4 are 36, and set it under the 37, as before, but do no: write the zeros by the side of it. I then subtract 36 from 37, and there remains 1. This of course is 100, but I do not mind it. I then bring down the 5 by the side of the 1, which makes 15, or rather 150, but I call it 15. Then I say, 4 into 15, 3 times, (this is 30, but I write only the 3;) I write the 3 by the side of the 9. Then I say, 3 times 4' is 12, which I write under the 15, and subtract it from 15, and there remains 3 (which is in fact 30.) By the side of 3 I bring down the 6, which makes 36.' Then I say 4 into 36, 9 times, which I write in the quotient, by the side of the 93, and it makes 939. The first 9 is now in the hundreds' place, and the 3 in the ten's place, as they ought to be. If this operation be compared with the last, it will be found in substance exactly like it. All the difference is, that in the last the figures are set down only when they are to be used.
A man employed a number of workmen, and gave them 27 dollars a month each ; at the expiration of one month, it took 10,125 dollars to pay them. How many men were there ?
It is evident that to find the number of men we must find how many times 27 dollars is contained in 10,125 dollars.
This may be done in the same manner as we did the last, though it is attended with rather more difficulty, because the divisor consists of two figures.
Operation. Dividend 10,125 (27 divisor
300 = the number of times 27 is contained 2,025
in 8,100 1,890 70 do.
1,890 5 do.
135 135 135 375 do.
I observe that there are not so many as 27 thousands, so I conclude that the divisor is not contained 1000 times in the dividend ; I therefore take the three left hand figures, neglecting the other two for the present. The three first are 101 ; (properly 10,100, but I notice only 101 ;) I seek how many times 27 is contained in 101, and find between 3 and 4 times. I put 3 in the quotient, which, when the work is done, must be 3 hundred, because 101 is 101 hundreds, but disregarding this circumstance, I find how much 3 times 27 is, and write it under 101. 3 times 27 are 81 ; this subtracted from 101, leaves 20. By the side of 20 I bring down 2, the next figure of the dividend which was not used. This makes 202, for the next partial dividend, I seek how many times 27 is contained in this. I find 7 times. I write 7 in the quotient. 7 times 27 are 159, which I subtract from 202, and find a remainder 13. By the side of 13 I bring down 5, the other figure of the dividend, which makes 135 for the last partial dividend. I find 27 is contained 5 times in this. I write 5 in the quotient. 5 times 27 are 135. There is no remainder, therefore the division is completed. Ans. 375 men.
The operation in the above example is precisely the same, as in those which precede it; but it is more difficult to discover how many times the divisor is contained in the partial dividends. When the divisor is still larger, the difficulty is increased. I shall next show how this difficulty may be obviated.
In 31,755 days, how many years, allowing 365 days to the
It is evident, that as niany times as 365 is contained in 31,755, so many years there will be.