Competition problem — Number theory

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Number theory Difficulty 5.0 AIME, harder Prove it Philippines

Problem:

Let $a$ and $b$ be integers for which $\frac{a}{2} + \frac{b}{1009} = \frac{1}{2018}$ . Find the smallest possible value of $|a b|$ .

Solution

Solution:

Clear denominators to write this as $1009 a + 2 b = 1$ . Clearly, $a = 1$ , $b = -504$ is a solution, and so our solutions are of the form $a = 1 + 2k$ , $b = -504 - 1009k$ . Now, clearly $|a| \geq 1$ , and $|b| \geq 504$ , so $|a b| \geq 504$ , and equality is attained when $a = 1$ and $b = -504$ .

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