Remainder when $333^{333}$ is divided by $7$

Question

Find the Remainder when $333^{333}$ is divided by $7$

I think I have to find $333^{333}\equiv r \pmod7$ where $r(\ge0)$ is the remainder but how do I get in that form

You should include what you know (or think you know) it is, and what you know about its order. — JMoravitz, Jun 28 '18 at 04:24
@John757 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work — user, Aug 05 '18 at 21:00

score 3 · Answer 1 · answered Jun 28 '18 at 04:24

3

Note that by FLT

$$333^6\equiv 1 \mod 7$$

thus

$$333^{333}\equiv 333^{3}\equiv 4^{3} \mod 7$$

answered Jun 28 '18 at 04:24

user

score 2 · Answer 2 · answered Jun 28 '18 at 04:21

2

$$333\equiv2^2\pmod7$$

and$$333\equiv3\pmod{\phi(7)}$$

$$\implies333^{333}\equiv(2^2)^3\equiv?\pmod7$$

answered Jun 28 '18 at 04:21

lab bhattacharjee

score 2 · Answer 3 · answered Jun 28 '18 at 04:31

2

$$333^{333}=(336-3)^{333}\equiv(-3)^{333}=-27^{111}\equiv-(-1)^{111}=1.$$

answered Jun 28 '18 at 04:31

Michael Rozenberg

3 Answers3