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.

See https://math.stackexchange.com/questions/679842/find-the-last-two-digits-of-345 and the related ones — lab bhattacharjee, Jun 25 '20 at 13:49

score 3 · Answer 1 · answered Jun 25 '20 at 13:48

3

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

answered Jun 25 '20 at 13:48

J. W. Tanner

score 0 · Answer 2 · answered Jun 25 '20 at 13:47

0

Using it mod 10 :

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

and mod 100 more brutally

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

answered Jun 25 '20 at 13:47

EDX

1

Use \equiv for $\equiv$ (that is what you mean, right?) – Andrew Chin Jun 25 '20 at 20:09
Yes equality stand because i'm talking about equivalence classes – EDX Jun 26 '20 at 09:04

2 Answers2