HMMT February, 2024

Number theory Difficulty 5.8 AIME, harder Prove it United States

Problem:

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 $x$ .

Solution

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