Consider the summation $\sum_{k=0}^n20^k$. Do the last digits of this always repeat? For example, with $n=54$ the summation is
$$\sum_{k=0}^{54} 20^k \\= 18\,962\,524\,746\,823\,141\,052\,631\,578\,947\,368\,421\,052\,631\,578\,947\,368\,421\,052\,631\,578\,947\,368\,421$$
Where the last 54 digits are 052 631 578 947 368 421 repeated 3 times. (I also plugged in higher values and got similar results)
Is there a proof of this result?
$\sum\limits_{k=0}^{54}20^k$ is a geometric series, so it's $\dfrac{20^{55}-1}{19}$.
$20^{55}=2^{55}\times10^{55}$, so it concludes with a long string of $0$s,
and $20^{55}-1$ concludes with a long string of $9$s.
Picture dividing $20^{55}-1$ by $19$ using long division.
Once the long string of $9$s starts, there will be a particular remainder,
which is a non-negative integer less than $19$.
Bring down the next $9$ and divide by $19$ to get another remainder.
Since there are only finitely many possible remainders,
eventually we will encounter a remainder that we have encountered before,
and at that point the cycle will repeat, until we run out of $9$s to bring down
and we are done dividing.