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2. 45616,875 – ,625=4561,6875 - ,0625 = 72987 72987

By simply changing the place of the comma in both numbers, or conceiving it to be changed, one place to the left, which (165) does not affect the quotient, we here recognize ,6875 and ,0625 as the numerators, 11 and 1, of their equivalents. Then, beginning on the left of ,6875, we multiply by 16, saying, 6 times 1 is 6, and 11 is 17, seven and go 1; then 6 times 6 is 36, and 1 is 37, and 1, on the right of 6, is 38; eight and go 3, and so on. 3. 136581,25 • ,9375 2185300 ; 15=

437060 = 3=145686,66, &c. Without regard to the comma, we begin on the left of 8125, and say, 6 times 5 is 30, and 13 is 43; three and go 4: then, 6 times 6 is 36, and 4 is 40; and 5 is 45, and so on; taking care to multiply the result by 100, because the comma is 2 places to the right.

4. 23587,40625 • ,4375 = 53914,0714285 + To divide by a fraction is to multiply by its reciprocal. Therefore, to divide by te is to multiply by L=24.


23587,40625 } twice the number

6739,25892% of the number.

53914,07142 + 5. 358127,95625 = 68,75. Multiply both by u. Then 57300,473 - 11=5209,13390909, ad inf.

The student has seen (176) that, in dividing by 11, the primary alternate figures, first, third, fifth, &c. are remainders; and the secondary—that is, the second, fourth, sixth, &c.—are undecimal complements of the remainders; also, that when the sums of primary and secondary are equal, the number is a multiple of 11. Hence, if the sum of the primary is the greater, the difference is the true remainder ; but, if that of the secondary is the greater, the undecimal complement of the difference is the remainder. Now, in the number 57300,473, as the sum of the primary is 14, and of the secondary 15, the complement of the difference 1, that is to say, 10 is the true remainder. Beginning with the right-hand


figure 3, we subtract the remainder, saying, 10 from 13, three. This we write as the right hand figure of the quotient. Then, before we again subtract, we add 1 to this 3, for the 10 previously added, and say, 4 from 7, three; which is the next figure of the quotient towards the left. Then, 3 from 4, one, which is the third, on the left of which we place the comma. Then, 1 from 10, nine; 10 from 10, nought; 1 from 3, two; 2 from 7, five; and 5 from 5, nought; which last we do not write. We thus obtain the quotient

5209,13310=5209,1339090 ad inf. This, which is merely the countermarch of the operation of multiplying by 11, (126,) we certainly do not give for its practical utility, but to accustom the student promptly to revert to previously acquired principles. 6. 77892,541875 : 812,5 =

1246,28067 -- 13 = 95,86774 + The scholar should, without again writing the numbers, divide thus : 13 in 124, nine times, and 7 over; writing 9 in the quotient : 13 in 76, five times, and 11 over; writing 5 and placing a comma on the right: then 13 in 112, eight times, and 8 over : 13 in 88, six times, and 10 over : 13 in 100, seven times, and 9 over : 13 in 96, seven times, and 5 over : 13 in 57, four times, and 5 over.

242. With an integral divisor, the quotient must (236) have as many decimals as the dividend. Now, if we divide the divisor, (165,) we multiply the quotient; therefore, (237) for every decimal in the divisor, we must move the comma in the quotient one place towards the right. Hence, there will ultimately remain in the quotient a number of decimals equal to the difference between the number in the dividend and the number in the divisor.

Again, the product of two decimal numbers has (234) as many decimals as both factors : therefore (60) the dividend has as many as the divisor and quotient. Hence, we have the following general rule : Separate by a comma a number of the right-hand figures of the quotient, equal to the number by which the decimal figures of the dividend exceed those of the divisor, and where the number of figures is not sufficient, supply the deficiency with ciphers, and prefix the comma. The previous methods are perhaps preferable.

1. ,0658 - 3256 =,000020208 +

2. ,654 : ,002191 =298,4938 + As the terminating figures 38 are much nearer to 40 than 30, if we required but 3 decimals, we should write 298,494 —. The sign - indicates that the quotient is somewhat too great.

3. 23+ ,71875=32; and 1=3,8125
4. 149,172 + ,496 =300,75
5. 63261 - 64,8 =-976,25
6. 52095,10875 6358,49 = 8,193 +
7. 3,141593 ,7854 =3,9999+
8. 149 - 15,36=9,7005208 +
9. 21,4 ,536 = 39,92537 +
10. ,2673 • 39=,00685384615 +
11. ,7103 - 8549 = ,0008386 —

12. 1,2193 = 136324=,000089440597 + 243. The work is often contracted by performing the subtraction, as we multiply each figure of the divisor, and writing, instead of the product, the remainder only. Thus, to divide 473804 by 899 :

899) 473804 (52739


31 Having found the quotient figure 5, by saying 9 in 47, we say, 5 times 9 is 45 : then, 45 from 8 being impossible, we say, 45 from 48, three, and write 3 under 8. Then, having added 4 tens to 8, we add 4 units to the next product: thus, 5 times 9 is 45, and 4 is 49. Then, 49 from 53, four, which we write under 3. Having added 5 tens, we add 5 units to the next product, thus : 5 times 8 is 40, and 5 is 45. Then, 45 from 47, two, which we write under 7. To the right of the remainder 243, we bring down the figure 0, and have 2430 for the next partial dividend.

With this we proceed in like manner, always adding as many tens as will enable us to subtract, and (85) 1 to the next product for each ten added to the upper figure.

244. When we only require the quotient within a unit, to contract the calculation, we suppress of the right-hand figures of the dividend as many, less one, as there are figures in the

divisor. Having exhausted this dividend, if there is no remainder, we place, on the right of the quotient, a cipher for each figure suppressed in the dividend. If there is a remainder, we consider it the new partial dividend. We then strike out, by a bar, as in cancelling, the right-hand figure of the divisor, always taking the remainder for a new dividend, and suppressing, at each division, the right-hand figure of the divisor, till the divisor is exhausted.

To find the quotient of 9968592476 = 52493, within a unit, we suppress the figures 2476 of the dividend, and divide 996859 by 52493, contracting the work as above, thus :

58ADB) 996859 (189903


4 We first divide by the whole divisor till we have exhausted the dividend, and have 18 for the quotient and 51985 for the remainder. We then bar the figure 3 in the divisor, and dividing by 5249, we have 9 for the quotient and 4744 for the remainder. Again, because 9 is nearly a unit of the next order to the left, in barring it, we add a unit to the next figure; and, dividing by 525, we have 9 for the quotient and 19 for the remainder. Barring again, we divide by 52, and have 0 for the quotient and 19 for remainder. Barring for the last time, we divide 19 by 5, and have 3 for the quotient and 4 for the remainder. Thus, then, we have obtained 189903 for the quotient, within a unit. The true quotient is 189903 14297

245. If, after barring the right-hand figure of the divisor, the partial dividend is too small to contain it, we place a cipher in the quotient, and again bar the right-hand figure of the divisor. If the dividend is still too small to contain the divisor, we place another cipher in the quotient, bar another figure of the divisor, and so on.

Also, when the division becomes exact, before the divisor is cxhausted, we place as many ciphers, less one, in the quotient as there are figures remaining in the divisor.

When the figure which we bar in the divisor, also when the left-hand figure of the part suppressed in the dividend, exceeds 5, we increase by a unit the next figure on the left.

524 93

To divide 1911740575899 by 546118, so that the quotient may be within a unit, we suppress the figures 75899 of the dividend; and, adding 1 to 5, the next figure on the left, we have 19117406, which we divide by 546118, thus :

516118) 19117406 (3500600




Having found the quotient 35 and the remainder 3276, as this does not contain the divisor 54612, after barring the figure 8, we place 0 in the quotient, and bar the next figure of the divisor; but, as 3276. does not contain 5461, we place another cipher, and bar the next figure. Then, as the divisor is contained exactly 6 times, we place 6 in the quotient, and, on the right, two ciphers for the two remaining figures, which we should have barred.

The true quotient being 3500599411113, the quotient, obtained by contraction, is not only within a unit, but within one-fifth of a unit.

246. If, after the suppression of the right-hand figures, the first dividend does not contain the divisor, we suppress, at once, as many as are necessary of the right-hand figures of the divisor, in order that it may be contained.

To have the quotient of 2349658 • 49786, within a unit, we first suppress the figures 9658 of the dividend, increasing, by a unit, those which remain. Then, as 235 does not contain 49786, we at once suppress the figures 786, and increase, by a unit, those which remain. We then divide 235 by 50, thus :

50) 235 (47


and have 47 for the quotient, which is within one-fifth of a unit; the true quotient being 4710788

247. With an integral divisor, the quotient is a number of units of the order of the right-hand figure of the dividend : consequently, we can, by the same method, find the quotient within a unit of any order of decimals ; namely, by assuming as many decimals on the right of the dividend as will reduce it to the required order.

Suppose we would have, within a ten-thousandth, the quo

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