geometric sequence number theory Problem: Find the remainder obtained upon dividing
$$1 + 11 + 11^2 + 11^3 + . . . + 11^{2019}$$
by 1000.
My go at it: Let the sum be $n$ and we know that $n = \frac{11^{2020}-1}{10}.$ By the chinese remainder theorem, we can just find the equation mod 125 and mod 8 and join them together later. After some calculation, we find that $n \equiv 0 \pmod{8}.$ As for mod 125, I got that $10n \equiv 75 \pmod{125}.$ However, I am having trouble dealing with the mod 125. Can anybody help? Thanks!
 A: Let consider
$$10x\equiv 11^{2020}-1 \pmod {10000} \iff 10x-1\equiv 11^{2020} \pmod {10000}$$
and following the hint given in the comments
$$11^{2020} = (10+1)^{2020} =\sum_{k=0}^{2020} \binom{2020}{k} 10^{2020-k}\equiv$$
$$\equiv  \binom{2020}{2020} +\binom{2020}{2019} 10+\binom{2020}{2018} 10^{2}+\binom{2020}{2017} 10^{3}=9201 \pmod {10000}$$
therefore
$$1 + 11 + 11^2 + 11^3 + . . . + 11^{2019}\equiv 920 \pmod{1000}$$
A: You need to add one more power of $5$ to take care of the factor of $5$ in the denominator.  Thus use $\bmod 625$.
Since $\phi(625)=500$, you may render $11^{2020}\equiv11^{20}$.  Since $11^5=161051\equiv1050+1$ and $1050$ is a multiple of $25$, the Binomial Theorem gives simply
$11^{20}=(11^5)^4\equiv(4×1050)+1\equiv4201\equiv451\bmod 625.$
Then you have $10x\equiv450\bmod 625$, and dividing by the common factor of $5$ gives $2x\equiv90\bmod 125$, which can be solved for a unique solution.
A: If you notice  in all terms the last digit is $1$ . The digits before varies from $0$ to $9$,there is 2020 terms, so we have:
$S_1=2020 \times 1=2020$
The sum of digits in each period from 0 to 9 is:
$s_2=\frac{9(9+1)}{2}=45$
so the sum of digits one before is:
$S_2=(\frac{2020}{10}=202)\times 45=9090$
Therefore:
$1+11+11^2+11^3+ . . . +11^{2019}\equiv (9090\times 10+2020=(92920) \ mod (1000)\equiv 920 \ mod (1000)$
