Maths Olympiad Prep

Library / /1218 of 1394

, 2024

Number theory Difficulty 5.8 AIME, harder Prove it United States

Problem:

Compute the sum of all two-digit positive integers xx such that for all three-digit (base 10) positive integers a b c\underline{a}\ \underline{b}\ \underline{c}, if a b c\underline{a}\ \underline{b}\ \underline{c} is a multiple of xx, then the three-digit (base 10) number b c a\underline{b}\ \underline{c}\ \underline{a} is also a multiple of xx.

Solution

Solution:

Note that abc0bca=a(1041)\overline{a b c 0} - \overline{b c a} = a\left(10^{4}-1\right) must also be a multiple of xx. Choosing a=1a=1 means that xx divides 103110^{3}-1, and this is clearly a necessary and sufficient condition. The only two-digit factors of 103110^{3}-1 are 2727 and 3737, so our answer is 27+37=6427+37=64.

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