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Common way.
10,125 (27
81

375 quotient.
202
189

135 135

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 oba viated.

In 31,755 days, how many years, allowing 365 days to the

year?

It is evident, that as many times as 365 is contained in 31,755, so many years there will be.

Operation.
Dividend 31755 (365 divisor
2920

87 quotient.
2555
2555

9, I say

I observe that 365 cannot be contained in 317, thercfore I must take the four left hand figures, viz. 3175. In order to discover how many times 365 is contained in this, I observe, that 365 is more than 300, and less than 400. I say 300 is contained in 3100, or simply 3 is contained in S1, 10 times, but 365 being greater than 300, cannot be contained in it more than 9 times. Indeed if it were contained more than 9 times, it must have been contained in 317, which is impossible. 400 is contained in 3100, (or 4 in 31) 7 times. This is the limit the other way, for 365 being less than 400, must be contained at least as many times. It is contained therefore 7, or 8, or 9 times. The most probable are 8 and 9. I try 9. But instead of multiplying the whole number 365 by

9 times 300 are 2700, or simply 9 times 3 are 27 ; then subtracting 2700 from 3170, there remains 470; I then say, 9 times 60 is 540, or simply 9 times 6 is 54, which being larger than 470, or 47, shows that the divisor is not contained 9 times. I next try 8 times, and say as before, 8 times 300 are 2400, which subtracted from 3170, leaves 770, then 8 times 60 are 480, which not being so large as 770, shows that the divisor is contained 8 times. I multiply the whole divisor by 8 (which is in fact 80,) the product is 2920. This subtracted from 3175 leaves 255. I then bring down the other 5, which makes the next partial dividend 2555. Now trying as before, I find that 3 is contained 8 times in 25, and I is contained 6 times. The limits are 6 and 8. It is probable that 7 is right. I multiply 365 by 7, and it makes 2555, which is exactly the number that I want. If I had wished to try 8, I should have said 8 times 3 are 24, which taken from 25 leaves 1. Then supposing 1 to be placed before the next figure, which is 5, it makes 15. 6 is not contained 8 times in 15, therefore 365 cannot be contained 8 times in 2555. The answer is 87 years.

The method of trying the first figure of the divisor into the first figure, or the first two figures of the partial divideni,

generally enables us to tell what the quotient figure must
be, within two or three, and it will always furnish the limits.
Then if we try the second figure, we shall always make the
limits smaller ; if any doubt then remains, which will not
often be the case, we may try the tiird, and so on.
Divide 436940074 by 64237.

Operation.
Dividend 436940074 (64237 divisor.
385422*

6802 quotient.
.515180
,513896*

128474 128474*

Proof 436940074 In this example I seek how many times 6, the first figure of the divisor, is contained in 43, the first two figures on the left of the dividend ; I find 7 times, and 7 is contained 6 times. The limits are 6 and 7. 7 times ù are 42, and 42 from 43 leaves 1, which I suppose placed by the side of 6; this makes 16. But 4, the second figure of the divisor, is not contained 7 times in 16, therefore 6 will be the first figure of the quotient.

it is easy to see that this must be 6000, when the division is completed ; because there being five figures in the divisor, and the first figure of the divisor being larger than the first figure of the dividend, we are obliged to take the six first figures of the dividend for the first partial dividend ; and the dividend containing nine figures, the right hand figure of this partial dividend, is in the thousands' place. I write 6 in the quotient, and multiply the divisor by it, and write the result under the dividend, so that the first figure on the right hand may stand under the sixth figure of the dividend, counted from the left, or under the place of thousands. This product, subtracted from the dividend as it stands, leaves a remainder 51518; by the side of this I bring down the next figure of the dividend, which is 0, and the second partial dividend is 515180. Trying as before with the 6, and then with the 4, into the first figures of this partial dividend, I find the divisor is contained in it 8 (800) times. I write 8 in the quotient, then multiplying and subtracting as before, I find a remainder 1284. I bring down the next figure of the dividend, which gives 12347 for the next partial dividend. I find that the divisor is not contained in this at all. I put 0 in the quotient, so that the other figures may stand in their proper places, when the division is completed. Then I bring down the next figure of the dividend, which gives for a partial dividend, 128474. The divisor is contained twice in this. Muitiplying and subtracting as before, I find no re mainder. The division therefore is completed.

Proof. It was observed in the commencement of this Art. that division is proved by multiplying the divisor by the quotient. This is always done during the operation. In the last example, the divisor was first multiplied by 6 (6000,) and then by 8 (800,) and then by 2 ; we have only to add these numbers together in the order they stand in, and if the work is right, this sum will be the dividend. The asterisms show the numbers to be added.

From the above examples we derive the following general rule for division : Place the divisor at the right of the dividend, scparate them by a mark, and draw a line under the divisor, to separate it from the quotient. Take as many figures on the left of the dividend as are necessary to contain the divisor once or morc. Seek how many times the first fig. ure of the divisor is contained in the first, or two first figures of these, then increasing the first figure of the divisor by one, scek hoo many times that is contained in the same figure or figures. Take the figure contained within these limits, which appears the most probable

, and multiply the two left hand figures of the divisor by it; if that is not sufficient to determine, multiply the third, and so on. When the first figure of the quotient is discovered, multiply the divisor by it, and subtract the product from the partial dividend. Then write the next figure of the dividend by the side of the remainder. This is the next partial dividenil. Scek as before how

many times the divisor is contained in this, and place the result in the quotient, at the right of the other quotient figure, then multiply and subtract, as before; and so on, until all the figures of the dividend have been used. If it happens that any partial dividend is not so large as the divisor, a zero must be put in the quotient, and the next figure of the dividend written at the right of the partial dividend.

Note. If the remainder at any time should exceed the

divisor, the quotient figure must be increased, and the multiplication and subtraction must be performed again. If the product of the divisor, by any quotient figure, should be larger than the partial dividend, the quotient figure must be diminished,

X

Short Division.

When the divisor is a small number, the operation of division may be much abridged, by performing the multiplication and subtraction in the mind without writing the results. In this case it is usual to write the quotient under the dividend. This method is called short division.

A man purchased a quantity of flour for 3045 dollurs, at 7 dollars a barrel. How many barrels were there? Long Division.

Short Division.
3045 (7

3045 (7
28
435

435
24
21

35
35

In short division, I say 7 into 30, 4 times ; I write 4 underneath ; then I

say 4 times or are 28, which taken from 30 leaves 2. I suppose the 2 written at the left of 4, which makes 24; then 7 into 24, 3 times, writing 3 underneath, I say 3 times 7 are 21, which taken from 24 leaves 3. I suppose the 3 written at the left of 5, which makes 35; then 7 in 35, 5 times exactly ; I write 5 underneath, and the division is completed.

If the work in the short and long be compared together, they will be found to be exactly alike, except in the short it is not written down.

ne X. How many yards of cloth, at 6 dollars a yard, may be bought for 45 dollars ?

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