How many three-digit numbers are there in which one of the digits is the sum of
the other two?
Case 1. Those which contain one 0, such as 404. (None can contain more than
one 0). The other two digits are the same. We can choose that non-zero digit 9
ways. The non-zero digit must come first. So for each of those there are
2! = 2 ways the other two digits can be arranged. So Case 1 contains 2∙9 = 18
ways.
Case 2. Those which have two digits alike.
They can only be like 112, 224, 336, 448. Each of those 4 can be arranged in
3!/2! = 6/2 = 3 ways. That's 12 for Case 2.
Case 3. Those which have no digits alike. We list the ways where the digits
are ascending.
123, 134, 145, 156, 167, 178, 189
235, 246, 257, 268, 279
347, 358, 369
459
That's 16. Each can be arranged 3!=6 ways. 16∙3! = 16∙6 = 96 ways for Case 3.
Total for all three cases = 18+12+96 = 126.
Edwin
