User:RandomUndergrad/Sandbox/Cyclic Permutations Preserve Divisibility

Theorem
Let $b>1$ be an integer.

Suppose $d$ is a strictly positive integer where $d\;\vert\;b^k-1$ for some strictly positive integer $k$.

Let $n$ be a $K$-Digit multiple of $d$ when expressed in base $b$, where $K$ is a multiple of $k$.

Let $m$ be an integer formed by cyclically permuting the digits of $n$.

Then $m$ is divisible by $d$.

Proof
Let $b>1$ be an integer.

Suppose $d$ is a strictly positive integer where $d\;\vert\;b^k-1$ for some strictly positive integer $k$.

Let $n$ be a $K$-Digit multiple of $d$ when expressed in base $b$, where $K$ is a multiple of $k$.

It suffices to show that the divisibility is preserved when digits of $n$ is shifted by 1 digit to the left,

i.e. $n=[n_1n_2\dots n_K]_b\mapsto[n_2n_3\dots n_Kn_1]_b$,

since this permutation generates the group of cyclic permutations on digits of $n$.

So we can let $m=[n_2n_3\dots n_Kn_1]_b$.

So $m$ is also divisible by $d$.

Examples

 * Both Cyclic Permutations of 5-Digit Multiples of 41 and Cyclic Permutation of 3-Digit Multiple of 37 are special cases of the theorem above since
 * $10^5-1=99999=3^2\times 41\times 271\text{ and }10^3-1=999=3^3\times 37$.
 * Here we take $b=10, d=41, k=K=5$ and $b=10, d=37, k=K=3$.


 * Since $10^{13}-1=9 999 999 999 999=3^2\times 53\times79\times265371653$,
 * the divisibility of both $53$ and $79$ are preserved under cyclic permutations for $13k$-digit integers.