<s>[INST] Finding remainder when $10^{10}+10^{10^2}+.........+10^{10^{10}}$ is divided by $7$. I have found a new problem which asks:

Find the remainder when $10^{10}+10^{10^2}+.........+10^{10^{10}}$ is divided by $7$.

I am thinking to find the remainder using Fermet's theorem, but I think I am unable to do it.
Please help.
 [/INST] Yes, You can use Fermat's theorem here. 
By Fermat's theorem, $10^6\equiv 1 \pmod 7$.
Hence, $10^{6m}\equiv 1 \pmod 7$ for all $m$.
Now, $$10\equiv 4 \pmod 6, 10^2\equiv 40 \equiv 4 \pmod 6$$
By induction, $10^n \equiv 4 \pmod 6$ for all $n$
Thus, $10^n=6m+4$
and $$10^{10n}=10^{6m}\times10^4\equiv 10^4 \pmod 7 \equiv 4 \pmod 7$$
Consequently,
$$10^{10}+10^{10^2}+..........+10^{10^{10}} \equiv 4 \times 10 \pmod 7\equiv 5 \pmod 7$$
So, remainder is $5$.
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