Find a nonagonal number that is the sum of three 3-digit prime numbers which together contain the nine nonzero digits once each.

EDIT

I'm adding this: 9-gonal (or enneagonal or nonagonal) numbers: n(7n-5)/2

0, 1, 9, 24, 46, 75, 111, 154, 204, 261, 325, 396, 474, 559, 651, 750, 856, 969, 1089, 1216, 1350, 1491, 1639, 1794, 1956, 2125, 2301, 2484, 2674, 2871, 3075, 3286, 3504, 3729, 3961, 4200, 4446, 4699, 4959, 5226, 5500, 5781, 6069, 6364

what do you mean by: contain the nine nonzero digits once each? ... they are not allowed to be a 0 in there...and the 3 primes must contain all digits, if you scramble them together?

ReplyDeleteUse the nonzero digits (1, 2,...,9) to form 3 primes (with three digits each), that equal n*(7n-5)/2 for some (integer) n.

ReplyDeleteHint: 149 + 257 + 683 = 1089 is the only solution.

ReplyDeleteCould you tell me why?