72. In the last example a number was taken which contains an exact number of thirteens. But this does not happen with every number. Take, for example, 159. Follow the process of (70), and it will appear that after having subtracted 13 twelve times, there remains 3, from which 13 cannot be subtracted. We may say then that 159 contains twelve thirteens and 3 over; or that 159, when divided by 13, gives a quotient 12, and a remainder 3. If we use signs,
159 = 13 × 12 + 3.
EXERCISES.
| 146 | = | 24 × 6 + 2, or 146 contains six twenty-fours and 2 over. |
| 146 | = | 6 × 24 + 2, or 146 contains twenty-four sixes and 2 over. |
| 300 | = | 42 × 7 + 6, or 300 contains seven forty-twos and 6 over. |
| 39624 | = | 7277 × 5 + 3239. |
73. If a contain b q times with a remainder r, a must be greater than bq by r; that is,
a = bq + r.
If there be no remainder, a = bq. Here a is the dividend, b the divisor, q the quotient, and r the remainder. In order to say that a contains b q times, we write,
a/b = q, or a : b = q,
which in old books is often found written thus:
a ÷ b = q.