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?
229, 361, 493, 625 is one solution.
They're in AP with d=132.
229 is prime, so number of factors=2.
361 = 19^2 so f=3.
493 = 17*29 so f=4.
625 = 5^4 so f=5.
I decided to look at the factor sequence 2,3,4,5 because there were only 2 candidates for f=5. 256 didn't work because that made the 2nd term a factor of 4, so the common difference was even making the first term even. A couple of checks with candidates using 625 gave a solution. No idea if there are additional solutions
Posted by xdog
on 2008-03-23 12:51:04