Sunday, February 13, 2011

Nonagonal Numbers

http://mathworld.wolfram.com/NonagonalNumber.html


A nonagonal number is a number of the form n*(7n-5)/2


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
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3 comments:

  1. nuitFebruary 13, 2011 at 4:54 PM

    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?

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  2. BenVitaleFebruary 14, 2011 at 12:05 PM

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

    ReplyDelete
  3. BenVitaleFebruary 19, 2011 at 12:14 PM

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

    Could you tell me why?

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