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Problem 130

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Composites with prime repunit property

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 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.

You are given that for all primes, p > 5, that p − 1 is divisible by A(p). For example, when p = 41, A(41) = 5, and 40 is divisible by 5.

However, there are rare composite values for which this is also true; the first five examples being 91, 259, 451, 481, and 703.

Find the sum of the first twenty-five composite values of n for which
GCD(n, 10) = 1 and n − 1 is divisible by A(n).

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满足素数循环单位数性质的合数

只包含数字1的数被称为循环单位数，我们定义R(k)是长为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，并且记A(n)是这些k中最小的一个。例如，A(7) = 6，而A(41) = 5。

已知对于素数p > 5，p − 1能够被A(p)整除。例如，当p = 41时，A(41) = 5，而40能够被5整除。

然而，有很少的一部分合数也满足这条性质，前5个这样的数分别是91，259，451，481以及703。

找出前25个合数n满足
GCD(n, 10) = 1且n − 1能够被A(n)整除，并求它们的和。

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