# Remainder of 823^823 divided by 11

• May 9th 2013, 10:32 AM
sirellwood
Remainder of 823^823 divided by 11
Hi, I am trying to work out what the remainder is when 823^823 is divided by 11. I know how to do it say if its 2^200 is divided by 7 where you find the remainder of 2^1 divided by 7 and 2^2 divided by 7 and find a recurring pattern and do it that way. That didnt seem to work with 823^823

Any ideas?

Thanks

• May 9th 2013, 10:57 AM
emakarov
Re: Remainder of 823^823 divided by 11
The method you describe works if you recall that 823 ≡ -2 (mod 11). (Change the base, not the exponent.)
• May 9th 2013, 11:33 AM
sirellwood
Re: Remainder of 823^823 divided by 11
ok, sorry im just not that comfortable with these questions so cant see how I can apply that to use my method I described?
• May 9th 2013, 11:42 AM
emakarov
Re: Remainder of 823^823 divided by 11
$\displaystyle \begin{array}{c|c|c|c|c|c|c}n&0&1&2&3&4&5\\ \hline (-2)^n\text{ mod }11 & 1&-2&4&-8&5&1\end{array}$

We see that the sequence of remainders when 823^n is divided by 11 is repeated after every 5 iterations. You just need to find the remainder corresponding to the 823th iteration.