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HMMT 十一月 2012 · 冲刺赛 · 第 19 题

HMMT November 2012 — Guts Round — Problem 19

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 11 ] Find the number of triples of nonnegative integers ( x, y, z ) such that 15 x + 21 y + 35 z = 525.
解析

英文解析

  1. [ 11 ]
    Answer: 21 First, note that 525 = 3 × 7 × 5 × 5. Then, taking the equation modulo 7 gives that
    ′ ′ ′ ′
    7 | x ; let x = 7 x for some nonnegative integer x . Similarly, we can write y = 5 y and z = 3 z for
    ′ ′
    some nonnegative integers y , z . Then, after substitution and division of both sides by 105, the given
    ′ ′ ′
    equation is equivalent to x + y + z = 5. This is the same as the problem of placing 2 dividers among
    ( )
    5 balls, so is = 21.7 2