HMMT, 2014

Solution:

Let $f(n)$ denote the number of $n$-tuples $\left(a_{1}, \ldots, a_{n}\right)$ such that $0 \leq a_{1}, \ldots, a_{n} \leq 7$ and $5 \mid 2^{a_{1}}+\ldots+2^{a_{n}}$. To compute $f(n+1)$ from $f(n)$, we note that given any $n$-tuple $\left(a_{1}, \ldots, a_{n}\right)$ such that $0 \leq a_{1}, \ldots, a_{n} \leq 7$ and $5 \nmid 2^{a_{1}}+\ldots+2^{a_{n}}$, there are exactly two possible values for $a_{n+1}$ such that $0 \leq a_{n+1} \leq 7$ and $5 \mid 2^{a_{1}}+\ldots+2^{a_{n+1}}$, because $2^{n} \equiv 1,2,4,3,1,2,4,3 \pmod{5}$ for $n=0,1,2,3,4,5,6,7$ respectively.

Also, given any valid $(n+1)$-tuple $\left(a_{1}, \ldots, a_{n+1}\right)$, we can remove $a_{n+1}$ to get an $n$-tuple $\left(a_{1}, \ldots, a_{n}\right)$ such that $0 \leq a_{1}, \ldots, a_{n} \leq 7$ and $5 \nmid 2^{a_{1}}+\ldots+2^{a_{n}}$, so these are in bijection. There are a total of $8^{n}$ $n$-tuples, $f(n)$ of which satisfy $5 \mid 2^{a_{1}}+\ldots+2^{a_{n}}$, so there are $8^{n}-f(n)$ for which $5 \nmid 2^{a_{1}}+\ldots+2^{a_{n}}$. Therefore, $f(n+1)=2\left(8^{n}-f(n)\right)$.

We now have $f(1)=0$, $f(2)=2(8-0)=16$, $f(3)=2(64-16)=96$, $f(4)=2(512-96)=832$, $f(5)=2(4096-832)=6528$.

Answer: $6528$