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. |
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 | = | b | + | c | + | d |
| m | m | m | 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: