HMMT 二月 2011 · 冲刺赛 · 第 22 题

HMMT February 2011 — Guts Round — Problem 22

[ 12 ] Find the number of ordered triples ( a, b, c ) of pairwise distinct integers such that − 31 ≤ a, b, c ≤ 31 and a + b + c > 0. 3

解析

[ 12 ] Find the number of ordered triples ( a, b, c ) of pairwise distinct integers such that − 31 ≤ a, b, c ≤ 31 and a + b + c > 0. Answer: 117690 We will find the number of such triples with a < b < c . The answer to the original problem will then be six times what we will get. By symmetry, the number of triples ( a, b, c ) with a + b + c > 0 is equal to the number of those with a + b + c < 0. Our main step is thus to find the number of triples with sum 0. If b = 0, then a = − c , and there are 31 such triples. We will count the number of such triples with b > 0 since the number of those with b < 0 will be equal by symmetry. For all positive n such that 1 ≤ n ≤ 15, if a = − 2 n , there are n − 1 pairs ( b, c ) such that a + b + c = 0 and b > 0, and for all positive n such that 1 ≤ n ≤ 16, if a = − 2 n + 1, there are also n − 1 such pairs ( b, c ). In total, we have 1 + 1 + 2 + 2 + 3 + 3 + ... + 14 + 14 + 15 = 225 triples in the case b > 0 (and hence likewise for b < 0.) In total, there are 31 + 225 + 225 = 481 triples such that a < b < c and a + b + c = 0. Since there ( ) 63 are = 39711 triples ( a, b, c ) such that − 31 ≤ a < b < c ≤ 31, the number of triples with the 3 39711 − 481 additional restriction that a + b + c > 0 is = 19615. So the answer to the original problem is 2 19615 × 6 = 117690. 3