<s>[INST] Proof that sum and product of geometric series is divisible by 10 I need to prove that for $x=7+7^2+\cdots+7^{2016}$, $y=8+8^2+\cdots+8^{2016}$, $10\mid x+y$ and $10\mid xy$. I think that if $10\mid(m_1+m_2)$, where $m_1$ is the remainder of $\frac{x}{10}$ etc., than $10\mid x+y$, but I can't determine the remainder. I also tried determining calculating the last digit of the geometric series, but I can't get past the division.
 [/INST] $$
7 + 7^2 + 7^3 + 7^4 \equiv 0 \pmod {10}
$$
Therefore
\begin{align}
& \overbrace{7 + 7^2 + 7^3 + 7^4} + \overbrace{7^5 + 7^6 + 7^7 + 7^8} + \overbrace{7^9 + 7^{10} + 7^{11} + 7^{12}} + \cdots \\[10pt]
= {} & \Big(7 + 7^2 + 7^3 + 7^4\Big) + 7^4\Big(7 + 7^2 + 7^3 + 7^4\Big) + 7^8 \Big(7 + 7^2 + 7^3 + 7^4\Big) + \cdots \\[10pt]
= {} & (10\times\text{something}) + 7^4(10\times\text{something}) + 7^8 (10\times\text{something}) + \cdots
\end{align}
A similar thing works with $8$.
</s>