Four three-digit numbers are in arithmetic progression, and the number of factors of each is also in arithmetic progression. In fact that second arithmetic progression has a constant difference of 1, so that each successive number in the original arithmetic progression has one more factor than the number before it.

Note: These are not the prime factors of the numbers, but rather any factor, including the number itself and 1, so, for example, 46 has four factors (1, 2, 23 and 46), as does 8 (1, 2, 4 and 8).

What is the original arithmetic progression of three-digit numbers?

I found only one such sequence:

229, 361, 493, 625 -- with constant difference of 132

229, 2 factors: 1, 229
361, 3 factors: 1, 19, 361
493, 4 factors: 1, 17, 29, 493
625, 5 factors: 1, 5, 25, 125, 625

Posted by Dej Mar on 2008-03-23 21:34:02