Maths Olympiad Prep

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, 2024

Number theory Difficulty 5.2 AIME, harder Prove it United States

Problem:

Positive integers aa, bb, and cc have the property that aba^{b}, bcb^{c}, and cac^{a} end in 44, 22, and 99, respectively. Compute the minimum possible value of a+b+ca+b+c.

Solution

Solution:

This minimum is attained when (a,b,c)=(2,2,13)(a, b, c) = (2, 2, 13). To show that we cannot do better, observe that aa must be even, so cc ends in 33 or 77. If c13c \geq 13, since aa and bb are even, it's clear (2,2,13)(2, 2, 13) is optimal. Otherwise, c=3c = 3 or c=7c = 7, in which case bcb^{c} can end in 22 only when bb ends in 88. However, no eighth power ends in 44, so we would need b18b \geq 18 (and a2a \geq 2), which makes the sum 2+18+3=232 + 18 + 3 = 23 larger than 1717.

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