AIME 2013 Solutions (divisiblity) Problem 2
Find the number of five-digit positive integers, $n$, that satisfy the following conditions:
(a) the number $n$ is divisible by $5,$
(b) the first and last digits of $n$ are equal, and
(c) the sum of the digits of $n$ is divisible by $5.$
Obviously, if $n$ satisfies divisiblity by five and takes the form,
$$n = abcda$$ then $a=5$, deductively, 
$$n = 5bcd5$$ And:
$$10 + b + c + d \equiv 0 \pmod{5} \implies b + c + d \equiv -10 \equiv 0 \pmod{5}$$
What do I do next? HINTS are appreciated!

Big hint: Note that the digits $b$ and $c$ can be chosen freely, ($100$ choices total); and then, whatever the choices for $b$ and $c$, there are $2$ choices for $d$.  For instance if $b$ and $c$ are chosen to be $7$ and $6$ respectively, then $d$ could be $2$ or $7$ to make $b+c+d$ congruent to $0 \pmod{5}$.
