Find the congruence of $4^{578} \pmod 7$

Question

Find the congruence of $4^{578} \pmod 7$.

Can anyone calculate the congruence without using computer?

Thank you!

Congruence Modulo with large exponents – Martin Sleziak Oct 21 '12 at 07:49 — Martin Sleziak, Oct 21 '12 at 07:49
Note that $4^3=64\equiv 1\pmod 7$. – Hagen von Eitzen Oct 21 '12 at 07:50 — Hagen von Eitzen, Oct 21 '12 at 07:50

score 6 · Answer 1 · edited Mar 17 '16 at 01:39

6

Use the fact that $$4^{3} \equiv 1 \ (\text{mod 7})$$ along with if $a \equiv b \ (\text{mod} \: m)$ then $a^{n} \equiv b^{n} \ (\text{mod}\: m)$.

edited Mar 17 '16 at 01:39

Mike Pierce

answered Oct 21 '12 at 08:02

Chandrasekhar

Jill Clover · Accepted Answer · 2012-11-01T16:19:52.897

1

Using Fermat's Little theorem:

If p is a prime and a is an integer, then $a^{p-1}\equiv1$ (mod p), if p does not divide a.

$4^{6}\equiv1 (mod 7)$

Since $4^{578}=(4^{6})^{96}\cdot4^{2}$, we can conclude that$4^{578}\equiv1^{96}\cdot4^{2}(mod 7)$.

Hence$4^{578}\equiv2(mod 7)$.

edited Nov 01 '12 at 16:19

answered Oct 22 '12 at 17:40

Jill Clover

2 Answers2