Last two digits of $529^{10}$

Question

Trying to find out how to get the last two digits of $529^{10}$.

I'm having trouble finding a good mod to reduce the $529$ down. Thanks.

Answer 1

Using it mod 10 :

$$529^{10}=9^{10}=81^5=1\mod 10 $$

and mod 100 more brutally

$$529^{10}=29^{10}=41^5=81^2*41= 1 \mod 100 $$

Answer 2

If you want the last two digits, take mod $100$, and $(530-1)^{10}\equiv 1\bmod 100$.