Determine all possible triplet(s) (A, B, C) of four-digit decimal primes, with A < B < C, such that:

(i) A, B and C (in this order) are in arithmetic sequence, and:

(ii) Each of B and C is obtained by permuting the digits of A, and:

(iii) None of A, B and C can contain any leading zero.

(In reply to re: computer exploration (spoiler?) -- another list by Charlie)

Charlie, I have complete faith and confidence in your method.

But, I would request you to recheck your calculations, since I know of at least one valid triplet that satisfies all the given conditions. (This triplet has a common difference of 3330).  

Posted by K Sengupta on 2009-01-30 15:50:26