unit of 314, it follows, that each unit of the 3d remainder, 6, which is a part of 62, is also 5 times as great as each unit of 314. Hence the remainder 6 is the same as 5 times 6, or 30, units of the same kind as those of 314; but the 2d remainder 4, being a part of 314, and of the same order, should be added to 30, making 34, for the true remainder arising from dividing 314 by 35 or 5×7. Again, since each unit of 314 is 3 times as great as each unit of 944, it follows, that each unit of the 34 is also 3 times as great as each unit of 944. Hence the remainder 34 is the same as 3 times 34–102 units of the same kind as 944; but the 1st remainder, 2, being a part of 944, is of the same order; so that 102+2=104, is the true remainder required. From the foregoing operation and reasoning, we deduce the following RULE. Divide the given sum by one of the factors of the divisor, and that quotient by another factor, and so on, until all the factors have been used. The last quotient will be the quotient sought. It makes no difference in what order the factors are used. To obtain the true remainder, we must observe the following RULE. Multiply the last remainder by the divisor preceding the last, and add in the preceding remainder; multiply this sum by the next preceding divisor, and add in the next preceding remainder; so continue this reverse process until you have multiplied by all the divisors except the last. Ans. 70056 with 7856 remainder. 10. Divide 166168212890625 by 12890625.* 11. Divide 11963109376 by 109376. Ans. 12890625. Ans. 109376. CASE III. 29. When the divisor is a composite number. We have seen (ART. 21,) that, in multiplication, when the multiplier is a composite number, the product may be found by multiplying by the factors successively. Now, as division is the reverse process of multiplication, it is plain that when the divisor is a composite number, the quotient may be found by dividing by the factors successively. Divide 944 by 105=3×5×7. In this division, we find the different remainders in succession. Let us now seek the true remainder, or that remainder which would have been found, had OPERATION. 1st factor 3)944 2d factor 5)314 2=1st rem. 3d factor 7)62 4=2d rem. quotient 8 6=3d rem. we at once divided the 944 by 105. Since each unit of the 62 is 5 times as great as each *This question and its succeeding one are worthy of notice, since the terminal figures of the dividend, divisor, and quotient, are the same. unit of 314, it follows, that each unit of the 3d remainder, 6, which is a part of 62, is also 5 times as great as each unit of 314. Hence the remainder 6 is the same as 5 times 6, or 30, units of the same kind as those of 314; but the 2d remainder 4, being a part of 314, and of the same order, should be added to 30, making 34, for the true remainder arising from dividing 314 by 35 or 5×7. Again, since each unit of 314 is 3 times as great as each unit of 944, it follows, that each unit of the 34 is also 3 times as great as each unit of 944. Hence the remainder 34 is the same as 3 times 34-102 units of the same kind as 944; but the 1st remainder, 2, being a part of 944, is of the same order; so that 102+2=104, is the true remainder required. From the foregoing operation and reasoning, we deduce the following RULE. Divide the given sum by one of the factors of the divisor, and that quotient by another factor, and so on, until all the factors have been used. The last quotient will be the quotient sought. It makes no difference in what order the factors are used. To obtain the true remainder, we must observe the following RULE. Multiply the last remainder by the divisor preceding the last, and add in the preceding remainder; multiply this sum by the next preceding divisor, and add in the next preceding remainder; so continue this reverse process until multiplied by all the divisors except the last. you have How do you proceed when the divisor is a composite number? Does it make any difference which factor we first divide by? When there are several remainders, explain how the true remainder is obtained. EXAMPLES. 1. Divide 839 by 120. We will resolve 120 into the three factors, 4×5×6= 120. Now, proceding agreeably to the rule, we have the annexed operation. OPERATION. 4)839 5)209 3=first rem. Now, to obtain the true remainder, we have this 5x5+4=29. Again, 29 x 4+3=119. Had there been more than three factors, the operation would have been equally simple, but a little more lengthy. 2. Divide 8217 by 35=5×7. Ans. 234 with 27 rem. 3. Divide 33678 by 15=3×5. 4. Divide 9591 by 72=8x9. 5. Divide 10859 by 497 x7. Ans. 2245 with 3 rem. Ans. 133 with 15 rem. Ans. 221 with 30 rem. CASE IV. 30. When the divisor ends with one or more ciphers. We have seen (ART. 4,) that a number is multiplied by 10 by annexing a cipher; it is multiplied by 100 by annexing two ciphers; by 1000 by annexing three ciphers, &c. Conversely, a number is divided by 10 by cutting off one figure from the right; it is divided by 100 by cutting off two figures from the right, &c. EXAMPLES. 1. Divide 2475 by 20. Having cut off the 5 from the right of the dividend, and the 0 from the right of the divisor, which is, in effect, dividing both dividend and OPERATION. 210)24715 123 15 remainder. divisor by 10, we proceed to divide 247 by 2, (ART. 25.) We obtain 123 for a quotient and 1 for a remainder. This remainder is 1 ten, since it is a part of the 7 of the dividend which occupies the ten's place; annexing the 5 units, which was cut off, to the 1 ten which remained, we have 1 ten and 5 units, or 15 for the true remainder. NOTE. This case may be comprised under Case III., Art. 29. Thus, taking the preceding example, the divisor 20=2× 10. Dividing 2475 first by 10, which division is effected by cutting off the right-hand figure, 5, we have 247 for the first quotient, and 5 for the first remainder. Next, dividing 247 by 2, we find 123 for the quotient sought, and 1 for the second remainder. Now, by the rule under the case referred to, we find the true remainder to be 1 x 10+5=15. |