Let A={$\displaystyle n \in N$ | 5 divides n}

B={$\displaystyle n \in N$| both n and $\displaystyle \frac {n}{5}$ contain only odd digits}

and C={$\displaystyle n \in N$ | both n and $\displaystyle \frac {n}{5}$ contain exactly 2006 digits} where N

denotes the set of all natural numbers.

Find the number of elements in $\displaystyle A \cap B \cap C$