2023/24 AHSMC Part 1 Problem 12

Problem 12 of 16

Let $a$ , $b$ , $c$ be positive integers such that $11a + 8b + 5c = 319$ . The smallest possible value of $a+b+c$ is

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We have $11a + 8b + 5c = 319 \iff 11(a+b+c-29) = 3b + 6c$ . Thus $3b+6c$ is divisible by $11$ and by $3$ , hence by $33$ . The minimum of $3b+6c$ subject to this is $33$ (e.g. $b=1$ , $c=5$ ). Then the minimum of $a+b+c$ is $32$ , achieved at $a=26$ , $b=1$ , $c=5$ . The answer is (B).

University of Alberta

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