HMMT 十一月 2015 · 团队赛 · 第 8 题

HMMT November 2015 — Team Round — Problem 8

[ 6 ] Find any quadruple of positive integers ( a, b, c, d ) satisfying a + b + c = d and abc < 10 .

解析

[ 6 ] Find any quadruple of positive integers ( a, b, c, d ) satisfying a + b + c = d and abc < 10 . Proposed by: Sam Korsky Answer: ( a, b, c, d ) = (128 , 32 , 16 , 4) or ( a, b, c, d ) = (160 , 16 , 8 , 4) x It’s easy to guess that there are solutions such that a, b, c, d are in the form of n , where n is a rather small number. After a few attempts, we can see that we obtain simple equations when n = 2 or n = 3: t t t +1 t +2 for n = 2, the equation becomes in the form of 2 + 2 + 2 = 2 for some non-negative integer t t t t +1 t ; for n = 3, the equation becomes in the form of 3 + 3 + 3 = 3 for some non-negative integer t . In the first case, we hope that t is a multiple of two of 3 , 4 , 5, that t + 1 is a multiple of the last one, and that t + 2 is a multiple of 11. Therefore, t ≡ 15 , 20 , 24 (mod 60) and t ≡ 9 (mod 11). It’s easy to check that the only solution that satisfies the given inequality is the solution with t = 20, and ( a, b, c, d ) = (128 , 32 , 16 , 4). In the case where n = 3, we must have that t is a multiple of 60, which obviously doesn’t satisfy the inequality restriction. Remark: By programming, we find that the only two solutions are ( a, b, c, d ) = (128 , 32 , 16 , 4) and ( a, b, c, d ) = (160 , 16 , 8 , 4), with the the former being the intended solution.