Modular arithmetic $x \equiv 17^{12} \pmod {11}$

Question

I'm trying to evaluate

$x \equiv 17^{12} \pmod {11}$

using modular arithmetic, but I'm a bit lost. I'd really appreciate a step-by-step on how to do it. Thanks!

As a start you may replace $17$ by $6$ because $$17 \equiv 6\pmod{11}$$ — AgentS, Mar 18 '15 at 07:36

score 1 · Accepted Answer · answered Mar 18 '15 at 07:38

The idea is to make a table.

$17\equiv 6\mod 11$, so $17^2\equiv 36=33+3\equiv 3\mod 11$. Thus $17^4=(17^2)^2=3^2=9\mod 11$. You can proceed from here.

Or you can do Fermat's Little Theorem. $17^{11}\equiv 17\mod11$, so $17^{12}\equiv (17^2)\equiv 3\mod 11$

score 0 · Answer 2 · answered Mar 18 '15 at 07:40

0

by F.M's little theorem:$$17^{12} = (11+6)^{12} = 6^{12} = 36\cdot 6^{10} = 36\cdot 1 = (3\cdot 11+3) = 3 \pmod {11}$$

answered Mar 18 '15 at 07:40

DeepSea

score 0 · Answer 3 · answered Mar 18 '15 at 07:42

0

Note that $17^{11}\equiv 17\equiv 6\mod 11$ by Fermat's little theorem. Hence, $17^{12}\equiv 3\mod 11$.

answered Mar 18 '15 at 07:42

3 Answers3