HMMT 二月 2015 · 冲刺赛 · 第 12 题

HMMT February 2015 — Guts Round — Problem 12

[ 6 ] For integers a, b, c, d , let f ( a, b, c, d ) denote the number of ordered pairs of integers ( x, y ) ∈ 2 { 1 , 2 , 3 , 4 , 5 } such that ax + by and cx + dy are both divisible by 5. Find the sum of all possible values of f ( a, b, c, d ). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT FEBRUARY 2015, 21 FEBRUARY 2015 — GUTS ROUND Organization Team Team ID# 3 2

解析

[ 6 ] For integers a, b, c, d , let f ( a, b, c, d ) denote the number of ordered pairs of integers ( x, y ) ∈ 2 { 1 , 2 , 3 , 4 , 5 } such that ax + by and cx + dy are both divisible by 5. Find the sum of all possible values of f ( a, b, c, d ). Answer: 31 Standard linear algebra over the field F (the integers modulo 5). The dimension of 5 the solution set is at least 0 and at most 2, and any intermediate value can also be attained. So the 2 answer is 1 + 5 + 5 = 31. This also can be easily reformulated in more concrete equation/congruence-solving terms, especially since there are few variables/equations. 3 2