Competition problem — Number theory

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Number theory Difficulty 5.0 AIME, harder Prove it Austria

Let $a$ , $b$ , $c$ and $d$ be four integers such that
$7a + 8b = 14c + 28d.$
Prove that $a \cdot b$ is a multiple of $14$ .

Solution

We consider the equation modulo $2$ and modulo $7$ , respectively, and obtain
$\begin{aligned} a &\equiv 0 \pmod{2}, \\ b &\equiv 0 \pmod{7}. \end{aligned}$
We conclude that $a$ is even and $b$ is a multiple of $7$ . Therefore, $ab$ is divisible by $2 \cdot 7$ . $\square$

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