But if we resolve it into parts, as we perform the operation, it will be done as follows: Dividend 3756 (4 divisor 3600 900 = number that 4 is cont'd in 3600 156 do 120 120 9 36 30 do do - 3756 36 939 36 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 iş 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 tiines. 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. 3756 (4 36 939 quotient. 15 12 36 36 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 not 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 tens' 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 8,100 300 = the number of times 27 is con2,035 tained in 8,100 1,590 70 do. 5 do. 135 135 135 375 do. 10,125 1,890 Common way. 375 quotient. 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 l 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 hundred, 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 189, 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 is 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 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 31755 days, how many years, allowing 365 duys to the year? It is evident, that as many times as 365 is contained in 31755, so many years there will be. Operation. 2920 87 quotient. I observe that 365 cannot be contained in 317, therefore 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 31, 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 tirnes. The niost probable are 8 and 9. I try 9. But instead of multiplying the whole number 365 by 9, 1 say 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 51, 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 ihat 3 is contained 8 times in 25, and 4 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 dividend, 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 third, and so on. Divide 436940074 by 64237. Operation. 6802 quotient. ... 128474 Proof. 436940074 In this example I seek how many times 6, the first figure of the divisor, is contained in 43, the two first |