# Thread: remainders over divisions

1. ## remainders over divisions

Hey guys, if i had a question of say:

QUESTION: find the remainders when the following numbers are divided by 89.

i) $809$

ii) $800^8$

is there an easy was to work this out?

thanks!

2. First note that $-a\equiv b-a\pmod{b}$.

It is easy to see that 10 * 89 = 809 + 81. Move 81 to the left side and consider both sides modulo 89.

If you represent 800 as $a * 89 + b$, then $800^8=(a * 89 + b)^8$. By the Binomial theorem (or just by expanding) one sees that all terms have the factor 89 except for the last one, namely, $b^8$.