Integer which is Multiplied by Last Digit when moving Last Digit to First

Theorem

Let $N$ be a positive integer expressed in decimal notation in the form:

$N = \sqbrk {a_k a_{k - 1} a_{k - 2} \ldots a_2 a_1}_{10}$

Let $N$ be such that when you multiply it by $a_1$, you get:

$a_1 N = \sqbrk {a_1 a_k a_{k - 1} \ldots a_3 a_2}_{10}$

Then at least one such $N$ is equal to the recurring part of the fraction:

$q = \dfrac {a_1} {10 a_1 - 1}$

Let us consider:

$q = 0 \cdotp \dot a_k a_{k - 1} a_{k - 2} \ldots a_2 \dot a_1$

Let:

$a_1 q = 0 \cdotp \dot a_1 a_k a_{k - 1} \ldots a_3 \dot a_2$

Then:

$\ds 10 a_1 q$

$=$

$\ds a_1 \cdotp \dot a_k a_{k - 1} a_{k - 2} \ldots a_2 \dot a_1$

$\ds \leadsto \ \ $

$\ds 10 a_1 q - a_1$

$=$

$\ds 0 \cdotp \dot a_k a_{k - 1} a_{k - 2} \ldots a_2 \dot a_1$

$\ds $

$=$

$\ds q$

$\ds \leadsto \ \ $

$\ds q$

$=$

$\ds \dfrac {a_1} {10 a_1 - 1}$

$\blacksquare$