Maths Olympiad Prep

Library / /388 of 740

, 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+c1\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 cb+a1c \mid b+a-1. WLOG assume c>b>ac>b>a; since b+a1<2cb+a-1<2c we must have b+a1=cb+a-1=c. Substituting into ba+c1b \mid a+c-1 and ac+b1a \mid c+b-1 gives
b2a2a2b2 \begin{aligned} & b \mid 2a-2 \\ & a \mid 2b-2 \end{aligned}
Since 2a2<2b2a-2<2b we must either have a=1a=1 (implying a=ba=b, bad) or 2a2=ba4a6a=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.