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Problem 129

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Repunit divisibility

A number consisting entirely of ones is called a repunit. We shall define $R(k)$ to be a repunit of length $k$; for example, $R(6) = 111111$.

Given that $n$ is a positive integer and $\text{GCD}(n, 10) = 1$, it can be shown that there always exists a value, $k$, for which $R(k)$ is divisible by $n$, and let $A(n)$ be the least such value of $k$; for example, $A(7) = 6$ and $A(41) = 5$.

The least value of $n$ for which $A(n)$ first exceeds ten is $17$.

Find the least value of $n$ for which $A(n)$ first exceeds one-million.

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循环单位数整除性

只包含数字$1$的数被称为循环单位数。定义$R(k)$是长度为$k$的循环单位数，例如，$R(6) = 111111$。

如果$n$是一个整数，且$GCD(n, 10) = 1$，可以验证总存在$k$使得$R(k)$能够被$n$整除，并记$A(n)$为这些$k$中最小的一个。例如，$A(7) = 6$，$A(41) = 5$。

最小的、满足$A(n)$大于十的$n$为$17$。

求最小的、满足$A(n)$大于一百万的$n$。

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