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, 2019

Number theory Difficulty 5.0 AIME, harder Prove it United States

Problem:
Let $a$ , $b$ , $c$ be positive integers such that
$\frac{a}{77} + \frac{b}{91} + \frac{c}{143} = 1$
What is the smallest possible value of $a + b + c$ ?

Solution

Solution:
We need $13a + 11b + 7c = 1001$ , which implies $13(a + b + c - 77) = 2b + 6c$ . Then $2b + 6c$ must be divisible by both $2$ and $13$ , so it is minimized at $26$ (e.g. with $b = 10$ , $c = 1$ ). This gives $a + b + c = 79$ .

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