Competition problem — Number theory

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Number theory Difficulty 5.0 AIME Prove it United States

Problem:

Compute the sum of all positive integers $a \leq 26$ for which there exist integers $b$ and $c$ such that $a+23 b+15 c-2$ and $2 a+5 b+14 c-8$ are both multiples of $26$ .

Solution

Solution:

Assume $b$ and $c$ exist. Considering the two values modulo $13$ , we find
$\begin{cases}a+10 b+2 c \equiv 2 & (\bmod\ 13) \\ 2 a+5 b+c \equiv 8 & (\bmod\ 13)\end{cases}$
Subtracting twice the second equation from the first, we get $-3 a \equiv -14\ (\bmod\ 13)$ . So, we have $a \equiv 9$ $(\bmod\ 13)$ . Therefore we must either have $a=9$ or $a=22$ .

Moreover, both $a=9$ and $a=22$ yield solutions with $b=0$ and $c=3,16$ , depending on the value of $a$ . Thus the answer is $9+22=31$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.