Problem 7

First, we factor $a^{2014}+a^{2015}$ as $a^{2014}(1+a)$. If $a=5$ or $a=10$, then the factor $a^{2014}$ is a multiple of 5, so the original expression is divisible by 5. If $a=4$ or $a=9$, then the factor $(1+a)$ is a multiple of 5, so the original expression is divisible by 5. If $a=1,2,3,6,7,8$, then neither $a^{2014}$ nor $(1+a)$ is a multiple of 5. Since neither factor is a multiple of 5, which is a prime number, then the product $a^{2014}(1+a)$ is not divisible by 5. Therefore, there are four integers $a$ in the range $1 \leq a \leq 10$ for which $a^{2014}+a^{2015}$ is divisible by 5.