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 One More Factor, Please (Posted on 2008-03-23)
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?

 See The Solution Submitted by Charlie Rating: 3.3333 (3 votes)

 Subject Author Date Answer K Sengupta 2008-12-31 00:39:16 re: Solution / method ==>>MY WAY Ady TZIDON 2008-03-24 20:43:13 Solution / method ed bottemiller 2008-03-24 15:19:59 Solution Dej Mar 2008-03-23 21:34:02 one solution xdog 2008-03-23 12:51:04

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