Maths Olympiad Prep

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HMMT November, 2022

Number theory Difficulty 5.0 AIME, harder Prove it United States

Problem:
A triple of positive integers (a,b,c)(a, b, c) is tasty if lcm⁡(a,b,c)∣a+b+c−1\operatorname{lcm}(a, b, c) \mid a+b+c-1 and a<b<ca<b<c. Find the sum of a+b+ca+b+c across all tasty triples.

Solution

Solution:
The condition implies c∣b+a−1c \mid b+a-1. WLOG assume c>b>ac>b>a; since b+a−1<2cb+a-1<2c we must have b+a−1=cb+a-1=c. Substituting into b∣a+c−1b \mid a+c-1 and a∣c+b−1a \mid c+b-1 gives
b∣2a−2a∣2b−2 \begin{aligned} & b \mid 2a-2 \\ & a \mid 2b-2 \end{aligned}
Since 2a−2<2b2a-2<2b we must either have a=1a=1 (implying a=ba=b, bad) or 2a−2=b⟹a∣4a−6⟹a=2,3,62a-2=b \Longrightarrow a \mid 4a-6 \Longrightarrow a=2,3,6. If a=2a=2 then b=2b=2. Otherwise, if a=3a=3 we get (3,4,6)(3,4,6) and if a=6a=6 we get (6,10,15)(6,10,15), so answer is 13+31=4413+31=44.

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