Question

Find the remainder obtained, when the number ${10}^{10}+{10}^{\left({10}^2\right)}+.......+{10}^{\left({10}^{10}\right)}$ is divided by 7.

Answer 1

By Fermat's theorem $={10}^6\equiv 1$ mod 7
Hence, ${10}^{6m}=1mod7$ for all m.
Now $10\equiv 4mod6,{10}^2\equiv 40\equiv 4mod6$
By induction ${10}^n\equiv 4mod6foralln.$
Thus, ${10}^n=6m+4$
and ${10}^{10n}={10}^{6m}{.10}^4={10}^{4}mod7$
$\equiv 4mod\left(7\right)$ consequently
${10}^{10}+{10}^{{10}^2}+.....+{10}^{\left({10}^{10}\right)}\equiv 4\times 10mod7$
The remainder is 5