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Algebra Difficulty 5.1 AIME, harder Find the answer

For integers a,b,c,da, b, c, d, let f(a,b,c,d)f(a, b, c, d) denote the number of ordered pairs of integers (x,y){1,2,3,4,5}2(x, y) \in \{1,2,3,4,5\}^{2} such that ax+bya x+b y and cx+dyc x+d y are both divisible by 5. Find the sum of all possible values of f(a,b,c,d)f(a, b, c, d).

A number or a short expression. Spacing and $ signs are ignored.

Solution

Standard linear algebra over the field F5\mathbb{F}_{5} (the integers modulo 5). The dimension of the solution set is at least 0 and at most 2, and any intermediate value can also be attained. So the answer is 1+5+52=311+5+5^{2}=31. This also can be easily reformulated in more concrete equation/congruence-solving terms, especially since there are few variables/equations.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.