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_. 74. If I divide 156 into several parts, and find how often 13 is contained in each of them, it is plain that 156 contains 13 as often as all its parts together. For example, 156 is made up of 91, 39, and 26. Of these 91 contains 13 7 times, 39 contains 13 3 times, 26 contains 13 2 times; therefore 91 + 39 + 26 contains 13 7 + 3 + 2 times, or 12 times. Again, 156 is made up of 100, 50, and 6. Now 100 contains 13 7 times and 9 over, 50 contains 13 3 times and 11 over, 6 contains 13 0 times[9] and 6 over. [9] To speak always in the same way, instead of saying that 6 does not contain 13, I say that it contains it 0 times and 6 over, which is merely saying that 6 is 6 more than nothing. Therefore 100 + 50 + 6 contains 13 7 + 3 + 0 times and 9 + 11 + 6 over; or 156 contains 13 10 times and 26 over. But 26 is itself 2 thirteens; therefore 156 contains 10 thirteens and 2 thirteens, or 12 thirteens. 75. The result of the last article is expressed by saying, that if _a_ = _b_ + _c_ + _d_, then _a_/_m_ = _b_/_m_ + _c_/_m_ + _d_/_m_ 76. In the first example I did not take away 13 more than once at a time, in order that the method might be as simple as possible. But if I know what is twice 13, 3 times 13, &c., I can take away as many thirteens at a time as I please, if I take care to mark at each step how many I take away. For example, take away 13 ten times at once from 156, that is, take away 130, and afterwards take away 13 twice, or take away 26, and the process is as follows: 156 130 10 times 13. --- 26 26 2 times 13. --- 0 Therefore 156 contains 13 10 + 2, or 12 times. Again, to divide 3096 by 18.