| I. 1782000) | 6424700000 | (3605 | 1782) | 6424700 | (3605 |
| 5346000 | 5346 | ||||
| 10787000 | 10787 | ||||
| 10692000 | 10692 | ||||
| 9500000 | 9500 | ||||
| 8910000 | 8910 | ||||
| 590000 | 590000 | ||||
| II. 12300000) | 42176189300 | (3428 | 123) | 421761 | (3428 |
| 36900000 | 369 | ||||
| 52761893 | 527 | ||||
| 49200000 | 492 | ||||
| 35618930 | 356 | ||||
| 24600000 | 246 | ||||
| 110189300 | 1101 | ||||
| 98400000 | 984 | ||||
| 11789300 | 11789300 | ||||
The rule, then, is: Strike out as many figures[12] from the right of the dividend as there are ciphers at the right of the divisor. Strike out all the ciphers from the divisor, and divide in the usual way; but at the end of the process place on the right of the remainder all those figures which were struck out of the dividend.
84. EXERCISES.
| Dividend. | Divisor. | Quotient. | Remainder. |
|---|---|---|---|
| 9694 | 47 | 206 | 12 |
| 175618 | 3136 | 56 | 2 |
| 23796484 | 130000 | 183 | 6484 |
| 14002564 | 1871 | 7484 | 0 |
| 310314420 | 7878 | 39390 | 0 |
| 3939040647 | 6889 | 571787 | 4 |
| 22876792454961 | 43046721 | 531441 | 0 |
Shew that
| 100 × 100 × 100 - 43 × 43 × 43 | |||
| I. | ———————————— | = | 100 × 100 + 100 × 43 + 43 × 43. |
| 100 - 43 | |||
| 100 × 100 × 100 + 43 × 43 × 43 | |||
| II. | ———————————— | = | 100 × 100 - 100 × 43 + 43 × 43. |
| 100 + 43 | |||
| 76 × 76 + 2 × 76 × 52 + 52 × 52 | |||
| III. | ———————————— | = | 76 + 52. |
| 76 + 52 | |||
| 12 × 12 × 12 × 12 - 1 | |||
| IV. | 1 + 12 + 12 × 12 + 12 × 12 × 12 | = | ————————. |
| 12 - 1 | |||
What is the nearest number to 1376429 which can be divided by 36300 without remainder?—Answer, 1379400.
If 36 oxen can eat 216 acres of grass in one year, and if a sheep eat half as much as an ox, how long will it take 49 oxen and 136 sheep together to eat 17550 acres?—Answer, 25 years.
85. Take any two numbers, one of which divides the other without remainder; for example, 32 and 4. Multiply both these numbers by any other number; for example, 6. The products will be 192 and 24. Now, 192 contains 24 just as often as 32 contains 4. Suppose 6 baskets, each containing 32 pebbles, the whole number of which will be 192. Take 4 from one basket, time after time, until that basket is empty. It is plain that if, instead of taking 4 from that basket, I take 4 from each, the whole 6 will be emptied together: that is, 6 times 32 contains 6 times 4 just as often as 32 contains 4. The same reasoning applies to other numbers, and therefore we do not alter the quotient if we multiply the dividend and divisor by the same number.
86. Again, suppose that 200 is to be divided by 50. Divide both the dividend and divisor by the same number; for example, 5. Then, 200 is 5 times 40, and 50 is 5 times 10. But by (85), 40 divided by 10 gives the same quotient as 5 times 40 divided by 5 times 10, and therefore the quotient of two numbers is not altered by dividing both the dividend and divisor by the same number.