Modular Arithmetic with high exponent?

Question

HOW?How do I simplify this expression?

$276^{247} \mod 323$

Thank you so much.

See also How do I compute $a^b,\bmod c$ by hand? and other posts linked there. — Martin Sleziak, Nov 21 '17 at 06:08
Although it's a bit unfortunate that after getting some answers to your original question about $382^{295} \bmod 11$ you have changed it to $276^{247} \bmod 323$. Meta: Question edited to change meaning. — Martin Sleziak, Nov 21 '17 at 06:11

score 0 · Answer 1 · edited Nov 21 '17 at 06:07

0

You can use Fermat's little theorem.

It tells us that $a^{10}\equiv 1 \bmod 11$ if $a$ is not a multiple of $10$.

It follows that $382^{295}= 382^{290} 382^{5}\equiv 1^{29}382^{5}=382^5\bmod 11$

This last one is easy to work out.

edited Nov 21 '17 at 06:07

Martin Sleziak

answered Nov 01 '17 at 01:19

Asinomás

score 0 · Answer 2 · answered Nov 01 '17 at 01:20

$$382^{295} \mod 11 \equiv 8^{295} \mod 11$$

This is by the definitions in modular arithmetic

If you are familiar with Fermat's little theorem, then you will know that $$p^{q-1} \mod q \equiv 1$$ if q is a prime and p is relatively prime to q

Therefore, we can rewrite it as

$$8^{295} \mod 11 \equiv 8^{10 \cdot 29} \cdot 8^5 \mod 11 \equiv 1^{29} \cdot {-3}^5 \mod 11 \equiv -243 \mod 11 \equiv 10 \mod 11$$

2 Answers2