Integer which is Multiplied by Last Digit when moving Last Digit to First
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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:
- $q = \dfrac {a_1} {10 a_1 - 1}$
Proof
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:
\(\ds 10 a_1 q\) | \(=\) | \(\ds a_1 \cdotp \dot a_k a_{k - 1} a_{k - 2} \ldots a_2 \dot a_1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 10 a_1 q - a_1\) | \(=\) | \(\ds 0 \cdotp \dot a_k a_{k - 1} a_{k - 2} \ldots a_2 \dot a_1\) | |||||||||||
\(\ds \) | \(=\) | \(\ds q\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds q\) | \(=\) | \(\ds \dfrac {a_1} {10 a_1 - 1}\) |
$\blacksquare$