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 $A(n)$.