--description--

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 $A(n)$.

--hints--

compositeRepunit() should return 149253.

assert.strictEqual(compositeRepunit(), 149253);

--seed--

--seed-contents--

function compositeRepunit() {

  return true;
}

compositeRepunit();

--solutions--

// solution required