Find the remainder $4444^{4444}$ when divided by 9

Question

When a number is divisible by 9 the possible remainder are $0, 1, 2,3, 4,5,6,7,8$ we know that $0$ is not a possible answer. My friend told me the answer is $7$ but how

I am thinking of taking 4444 divide by 9 and that left a remainder of 7 so $4444 \cong 7$ mod 9 based on that I am guessing $4444^{4444} \cong 7^{4444}$ mod 9 so to me no matter how big the power is that will still be a remainder of 7 is that correct to think like that

You're on the right track. Have you learned any shortcuts to computing residues for large powers? (I'm thinking of one that involves expressing the exponent in binary.) — Sammy Black, Sep 28 '15 at 21:10
@user146269 Look at $7, 7^2, 7^3, 7^4, \ldots$ mod $9$ and see if you can find a pattern. — Erick Wong, Sep 29 '15 at 05:26

Zestylemonzi · Answer 1 · 2015-09-28T21:33:24.560

4

Hint: You've got the right idea! Notice now that $7 \cong -2 \pmod 9$ and $(-2)^3 \cong -8 \cong 1 \pmod 9$. Using this can you write $4444$ as a multiple of a certain (useful) number plus a remainder term?

Extra: An example for clarification. What's the remainder on dividing $33^{33}$ by $7$? We have: $$33 \cong 5 \pmod 7$$ so $$33^{33} \cong 5^{33} \pmod 7$$ But now we can notice that $5^3 \cong -1 \cong 6\pmod 7$ and therefore: $$33^{33} \cong 5^{33} \cong 5^{3 \times 11} \cong (5^3)^{11} \cong (-1)^{11} \cong -1 \cong 6\pmod 7$$ So $33^{33}$ has remainder $6$ on division by $7$. Can you try a similar thing with your example?

edited Sep 28 '15 at 21:33

answered Sep 28 '15 at 21:11

Zestylemonzi

4,103

Feel free to ask for more hints/clarification if you need it :) – Zestylemonzi Sep 28 '15 at 21:16
i do not quite follow it is because i know that $4444 \cong 7$ mod 9 that is why i was able to derive that. elaborate more in the hint – user146269 Sep 28 '15 at 21:19
I'll add a simpler example to my answer. – Zestylemonzi Sep 28 '15 at 21:20
after i follow your example i have (-2)^1111 mod 9 how do i simplify this – user146269 Sep 28 '15 at 21:47
Sounds good to me, now notice that $1111 = (3 \times 370) + 1$, so now....? – Zestylemonzi Sep 28 '15 at 21:53
i know that $7^4 \cong 7 \cong (-2)$ mod 9 so $7^{4444} \cong 7^{4 \dot 1111} \cong (-2)^{1111}$ then i don't see how this reduces to just $(-2)$ – user146269 Sep 28 '15 at 21:57
Let us continue this discussion in chat. – Zestylemonzi Sep 28 '15 at 21:57

score 1 · Answer 2 · edited Jun 22 '16 at 14:31

1

Using the Euler's theorem ( $a^{\phi(m)}\equiv 1 \pmod m$) you can write: $$ 4444^{4444}\equiv 4444^{6\cdot 740+4}\equiv 4444^4\equiv 7^4\equiv 25\equiv 7\pmod 9$$

edited Jun 22 '16 at 14:31

Martin Sleziak

53,687

answered Sep 29 '15 at 16:25

Domenico Vuono

2,672

yeah unfortunately i can not use it. A friend of mine was telling me about it. thanks though – user146269 Sep 29 '15 at 16:30
1

Please do not skip the step of checking if $a$ is coprime to $9$. – Element118 Jan 06 '16 at 23:03

score 0 · Answer 3 · answered Sep 28 '15 at 21:33

0

You are on the right track. You may now want to have a look at modular exponentiation.

answered Sep 28 '15 at 21:33

Roy Stark

139

this is not helping me getting anywhere – user146269 Sep 28 '15 at 21:34

Find the remainder $4444^{4444}$ when divided by 9

3 Answers3