HMMT 二月 2017 · 冲刺赛 · 第 4 题

HMMT February 2017 — Guts Round — Problem 4

[ 4 ] Find the number of ordered triples of nonnegative integers ( a, b, c ) that satisfy ( ab + 1)( bc + 1)( ca + 1) = 84 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . February 2017, February 18, 2017 — GUTS ROUND Organization Team Team ID#

解析

[ 4 ] Find the number of ordered triples of nonnegative integers ( a, b, c ) that satisfy ( ab + 1)( bc + 1)( ca + 1) = 84 . Proposed by: Evan Chen Answer: 12 The solutions are (0 , 1 , 83) and (1 , 2 , 3) up to permutation. First, we do the case where at least one of a, b, c is 0. WLOG, say a = 0 . Then we have 1 + bc = 84 = ⇒ bc = 83 . As 83 is prime, the only solution is (0 , 1 , 83) up to permutation. Otherwise, we claim that at least one of a, b, c is equal to 1. Otherwise, all are at least 2, so (1 + 3 ab )(1 + bc )(1 + ac ) ≥ 5 > 84 . So WLOG, set a = 1 . We now need ( b + 1)( c + 1)( bc + 1) = 84 . 2 Now, WLOG, say b ≤ c. If b = 1, then ( c + 1) = 42, which has no solution. If b ≥ 3, then 2 ( b + 1)( c + 1)( bc + 1) ≥ 4 · 10 = 160 > 84 . So we need b = 2 . Then we need ( c + 1)(2 c + 1) = 21 . Solving this gives c = 3, for the solution (1 , 2 , 3) . Therefore, the answer is 6 + 6 = 12 .