Find remainder when $777^{777}$ is divided by $16$

Question

Find remainder when $777^{777}$ is divided by $16$.

$777=48\times 16+9$. Then $777\equiv 9 \pmod{16}$.

Also by Fermat's theorem, $777^{16-1}\equiv 1 \pmod{16}$ i.e $777^{15}\equiv 1 \pmod{16}$.

Also $777=51\times 15+4$. Therefore,

$777^{777}=777^{51\times 15+4}={(777^{15})}^{51}\cdot777^4\equiv 1^{15}\cdot 9^4 \pmod{16}$ leading to $ 81\cdot81 \pmod{16} \equiv 1 \pmod{16}$.

But answer given for this question is $9$. Please suggest.

You can't use Fermat here - $16$ is not prime. You have to use Euler's theorem: https://en.wikipedia.org/wiki/Euler's_theorem — Wojowu, Jun 28 '15 at 13:04
Oops, you were faster, shall I mark you or delete my answer?: ) — Atvin, Jun 28 '15 at 13:05
@Wojowu Oh sorry, yes I was wrong. Please suggest solution for me. — math131, Jun 28 '15 at 13:05
http://math.stackexchange.com/questions/623008/determination-of-the-last-two-digits-of-777777 — Pratyush, Jun 28 '15 at 13:06

score 5 · Answer 1 · answered Jun 28 '15 at 13:09

5

Alternative to other answers: Note that $9^2=81\equiv 1\pmod{16}$, so $777^{777}\equiv 9^{777}=9\cdot(9^2)^{388}\equiv 9\pmod{16}$.

answered Jun 28 '15 at 13:09

Wojowu

27,526

1

Nice alternate solution, +1 :) – Atvin Jun 28 '15 at 13:11

Atvin · Answer 2 · 2015-06-28T13:10:38.557

1

Fermat's theorem will only work with primes.($16$ is not a prime)

But, it can be solved by the general formula, using Euler's theorem.

Since $gcd(9,16)=1, 9^{\phi(16)} \equiv 1 (mod 16)$. Since $16=2^4, \phi(16)=8$.

Therefore, $9^8 \equiv 1 (mod 16)$, and we have $777^{777} \equiv 9^{777} \equiv 9^9 \equiv 9*9^8 \equiv 9 (mod 16)$.

edited Jun 28 '15 at 13:10

answered Jun 28 '15 at 13:05

Atvin

3,512

$777^{777} \equiv 9^{777}$ please explain this step. – math131 Jun 28 '15 at 13:13
You already did this step in your solution, $777=48$x$16+9$, therefore, it is equivalent with $9$ modulo $16$ – Atvin Jun 28 '15 at 13:15
Ok, I understand – math131 Jun 28 '15 at 13:17
It doesn't matter, because $777$ and $9$ are in the same equivalent class, they count as the same number, when working modulo $16$, can't explain it any longer :) – Atvin Jun 28 '15 at 13:18
$9^{777}≡9^9$ I am unable to understand. Did you use 777=96x8+9 ? – math131 Jun 28 '15 at 13:28
$9^1,9^2,...,9^{16}$ are different classes modulo $16$, but it is easy to see, that, for example, $9^{17} \equiv 9^1$ modulo $16$, because $17 \equiv 1$ modulo $16$, you can use this method with the exponent too. – Atvin Jun 28 '15 at 13:37

score 1 · Answer 3 · answered Jun 28 '15 at 13:06

1

Since $9$ is coprime to $16$, you know that $$ 9^{\varphi(16)}\equiv 1\pmod{16} $$ where $\varphi$ is Euler's totient function. Your method is good, but $\phi(16)\ne 16-1$, so you got the answer wrong.

answered Jun 28 '15 at 13:06

egreg

244,946

Yes, I was wrong. – math131 Jun 28 '15 at 13:14

score 1 · Answer 4 · answered Jun 28 '15 at 13:11

1

Hint: $9^2~=~81~=~5\cdot16+1$.

answered Jun 28 '15 at 13:11

Lucian

49,312

Find remainder when $777^{777}$ is divided by $16$

4 Answers4