Prove that $100|11^{10} - 1$

Question

I want to prove divisibility using factoring, So i need to show that $11^{10}-1$ can be written as prime factors of 100.

This is what I've tried: $$11^{10}-1 $$ $$ (11^{5})^{2}-1$$ $$ (11^{5}-1)(11^{5}+1 )$$ $$10(11^{4}+11^{3}+11^{2}+11^1+1)(12)(11^4-11^3+11^2-11^1+1)$$ $$5*2^3*3[(11^4+11^2+1)+(11^3+11)]*[(11^4+11^2+1)-(11^3+11)]$$

I wanna write the $[(11^4+11^2+1)+(11^3+11)]*[(11^4+11^2+1)-(11^3+11)]$ as $5k$ with $k$ $\in\mathbb{Z}$.

I'd love any help/hint that would move me forward.

Answer 1

Hint: $$11\equiv 1 \mod 10$$ so $$11^{10}\equiv 1^{10}\equiv 1 \mod 10$$