Question 62
For a positive integer $n$, we consider all its divisors (including $1$ and itself). Suppose that $p\%$ of these divisors have their unit digit equal to $3$ (For example $n=117$, has six divisors, namely $1, 3, 9, 13, 39, 117$. Two of these divisors, namely $3$ and $13$, have unit digits equal to $3$. Hence for $n=117$, $p=33.33\cdots$). Find, when $n$ is any positive integer, the maximum possible value of $p$.