Competition problem — Number theory

Library / /674 of 841

Number theory Difficulty 5.4 AIME, harder Find the answer

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

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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

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Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.