HMMT 二月 2011 · 冲刺赛 · 第 22 题

英文解析

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.
    Answer: 117690
    We will find the number of such triples with a < b < c . The answer to the original problem will thenbe six times what we will get. By symmetry, the number of triples ( a, b, c ) with a + b + c > 0 is equalto the number of those with a + b + c < 0. Our main step is thus to find the number of triples withsum 0.
    If b = 0, then a = − c , and there are 31 such triples. We will count the number of such triples withb > 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 (andhence 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
    ( )
    are = 39711 triples ( a, b, c ) such that − 31 ≤ a < b < c ≤ 31, the number of triples with the 63
    39711 − 4813
    additional restriction that a + b + c > 0 is = 19615. So the answer to the original problem is
    19615 × 6 = 117690.2 3