<s>[INST] Number theory - congruence 
Let $n$ be an even integer not divisible by $10$, what digit is in the $10$s place for $n^{20}$ and the hundreds place for $n^{200}$, can you generalise this?

I know that for $n^{20}$ they end in $76$ and for $n^{200}$ they all end in $376$
I even went on too see that $n^{2000}$ always ends in $9376$.
I was given a clue that $76$ is the only number that is divisible by $4$ that gives $1$ when working $\bmod{25}$, I'm not sure how to relate that and generalise
When working $\bmod{100}$ for $n^{20}$ I noticed, $100 = 4 \times 25 = 2^2 \times 5^2$ , so we looked at a number divisible by $4$ that gave $1$ when working in $\bmod{25}$.
Similarly for $n^{200}$, $1000 = 2^3 \times 5^3$, for which $376$ is divisible by $8$ and gives $1$ when working with $\bmod{125}$?
Am I looking along the right lines?
Any help would be greatly appreciated.
Thanks!
 [/INST] Let us first do the case $n^{20} \pmod{100}$.
You have been advised to split this into two problems


*

*$n^{20} \pmod{4}$. Since $n$ is even, this yields $n^{20} \equiv 0 \pmod{4}$ here.

*$n^{20} \pmod{25}$. Since $n$ is not divisible by $5$, we have $(n, 25) = 1$. Since $\varphi(25) = 20$, we obtain  $n^{20} \equiv 1 \pmod{25}$.


(Here $\varphi$ is Euler's totient function.)
Now you solve the system of congruences
$$
\begin{cases}
x \equiv 0 \pmod{4}\\
x \equiv 1 \pmod{25}
\end{cases}
$$
which has the solution(s) $x \equiv 76 \pmod{100}$.
In the general case $k \ge 2$ you have $n^{2 \cdot 10^{k-1}} \pmod{10^{k}}$. Again, split it into two problems


*

*$n^{2 \cdot 10^{k-1}} \equiv 0 \pmod{2^k}$, as above.

*$n^{2 \cdot 10^{k-1}} \pmod{5^k}$. Since $n$ is not divisible by $5$, we have $(n, 5^k) = 1$. Since $\varphi(5^k) = 5^k - 5^{k-1} = 4 \cdot 5^{k-1}$, we obtain  $n^{2 \cdot 10^{k-1}} \equiv 1 \pmod{5^k}$.


Now you solve the system of congruences
$$
\begin{cases}
x \equiv 0 \pmod{2^k}\\
x \equiv 1 \pmod{5^k}.
\end{cases}
$$ 
</s>