Numbers whose Cyclic Permutations of 3-Digit Multiples are Multiples

Theorem
Let $n$ be a two-digit positive integer with the following property:


 * Let $m$ be a $3$-digit multiple of $n$.


 * Then any cyclic permutation of the digits of $m$ is also a multiple of $n$.

Then $n$ is either $27$ or $37$.

Proof
Let $m$ be a multiple of $n$ with $3$ digits.

Then we have:

Let us now cyclically permute the digits of $m$ by multiplying by $10$.

Then we have:

From the above, we see that:
 * $n$ is a divisor of a cyclic permutation of $m$


 * $n \divides \paren {10^3 - 1 }$
 * $n \divides \paren {10^3 - 1 }$

We now note that:
 * $10^3 - 1 = 37 \times 27 = 37 \times 3^3$

Upon inspection, we see that the only $2$-digit factors are $27$ and $37$.

Also See

 * Cyclic Permutations of 5-Digit Multiples of 41