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Clearly A, B, and C are digits between zero and eight. A takes any value from 1 to 8.* There are thus eight possibilities for the value of A. Each value of A will dictate a value N, from 0 to 7, for (B+C). For a given value of N, B can take any value from 0 to N, and thus can take N + 1 different values, each corresponding to a particular value of C. So there are N + 1 ways of achieving (B+C) = N. The total number of combinations is therefore add up [N+1] between N=0 and N=7 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = (8 + 1) x 8/2 = 9 x 4 = 36 And here are those 36 solutions: 800,701,710,611,620,602,521,512,530,503,422,431,413,440,404,332,323,341,314,350,305,233,242,224,215,251,206,260,143,134,152,125,116,161,107,170 *If A was zero, we wouldn't really have a three digit number as required.

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Q: How many three figure numbers have digits that add up to eight?
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