0

How do you solve $7^{60} \pmod{77}$?

For $5^{60} \pmod{77}$, it's equal to $1$ because I think the $\gcd(5,77) = 1$ and using the FLT we can find that $5^{(7-1)(11-1)} = 1 \pmod{7*11}$. But how do you get $56$ from $7^{60} \pmod{77}$?

Giani Kurniawan
  • 9
  • 2
    CRT with $7^{60}\pmod{11}$ and $7^{60}\equiv 0\pmod{7}$ – Joshua Wang Dec 16 '20 at 14:40
  • Note: $7^{10}\equiv1\bmod11$ – J. W. Tanner Dec 16 '20 at 14:50
  • See https://math.stackexchange.com/questions/3932977/remainder-of-division-when-p-is-not-prime/3932990#3932990 and https://math.stackexchange.com/questions/3789602/finding-the-last-2-digits-of-6513/3789621#3789621 – lab bhattacharjee Dec 16 '20 at 15:20
  • $7^{60}!\bmod 77 = 7 (\overbrace{7^{60}!/7 \bmod 11}^{\textstyle \color{#c00}{1/7}\ {\rm by} \ 7^{10}!\equiv 1!!!!!!!}) = 7(\color{#c00}8)\ $ by $\bmod 11!:\ \color{#c00}{\dfrac{1}7} \equiv\dfrac{12}{-4} \equiv -3 \equiv\color{#c00}8\ $ as here. We have hundreds of questions on this topic. Please search first. – Bill Dubuque Dec 16 '20 at 16:03

1 Answers1

1

First, $7^{60}\equiv 0\pmod{7}$. Then:

$$7^{60}\equiv \big(7^{10}\big)^{6}\equiv \big(7^{\phi(11)}\big)^{6}\equiv 1^{6}\equiv 1\pmod{11}$$

Then, $7^{60}\equiv 0\pmod{7}$ and $7^{60}\equiv 1\pmod{11}$, so $7^{60}$ is equivalent to a unique value $\pmod{77}$ by the Chinese Remainder Theorem. And since $56\equiv 0\pmod{7}, 1\pmod{11}$, we must have $\boxed{7^{60}\equiv 56\pmod{77}.}$

Joshua Wang
  • 6,103