Numbers whose Cyclic Permutations of 3-Digit Multiples are Multiples

From ProofWiki
Jump to navigation Jump to search

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:

\(\displaystyle n \times c\) \(=\) \(\displaystyle a_2 \times 10^2 + a_1 \times 10^1 + a_0\)


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

Then we have:

\(\displaystyle 10 \times n \times c\) \(=\) \(\displaystyle 10 \times \paren {a_2 \times 10^2 + a_1 \times 10^1 + a_0}\) multiply original number by $10$
\(\displaystyle \) \(=\) \(\displaystyle a_2 \times 10^3 + a_1 \times 10^2 + a_0 \times 10^1\)
\(\displaystyle \) \(=\) \(\displaystyle a_1 \times 10^2 + a_0 \times 10^1 + a_2 \times 10^0\) $10^3$ and $10^0 \equiv 1 \pmod {n}$


From the above, we see that:

$n$ is a divisor of a cyclic permutation of $m$

if and only if:

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


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 $37$.

$\blacksquare$


Also see


Sources