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 HeaRT OF VENUS (Posted on 2011-09-18)
Solve the following alphametic, given that two are primes and one is a composite:

SEVEN - THREE = FOUR

BONUS: Without the restriction of the number of composites and primes in the alphametic, how many different solutions are there?

 Submitted by Dej Mar No Rating Solution: (Hide) U V E F O T S R H N 0 1 2 3 4 5 6 7 8 9 SEVEN - THREE = FOUR 62129 - 58722 = 3407 62129 and 3407 are the two primes with 58722 as the composite. Without the restriction of the number of primes and composites there would be 38 possible solutions for the alphametic. Including the solution, those numbers are: SEVEN THREE FOUR 23439 15633 7806 23439 17633 5806 23938 16533 7405 23938 17533 6405 24349 16544 7805 24349 17544 6805 25758 16355 9403 25758 19355 6403 31519 24811 6708 31519 26811 4708 35159 26455 8704 35159 28455 6704 36061 27566 8495 36061 28566 7495 41517 32611 8906 41517 38611 2906 41918 35711 6207 41918 36711 5207 45157 36255 8902 45157 38255 6902 52728 43622 9106 52728 49622 3106 56368 47266 9102 56368 49266 7102 61219 53811 7408 61219 57811 3408 62129 53722 8407 62129 58722 3407 71315 62411 8904 71315 68411 2904 71814 62311 9503 71814 69311 2503 73135 64233 8902 73135 68233 4902 82526 73422 9104 82526 79422 3104 84346 75244 9102 84346 79244 5102 It might be interesting to note that two of the solutions (where SEVEN=36061) U=9, with the other 36 solutions U=0.

 Subject Author Date computer solution Charlie 2011-09-18 17:01:49 PARTIAL SOLUTION Ady TZIDON 2011-09-18 14:43:06 PARTIAL SOLUTION Ady TZIDON 2011-09-18 14:43:03

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