0

I am doing practice problems for an upcoming exam and am wondering what approach I might take to solve $53^{1069}$ mod 54. 1069 is a prime number, which means I can't factor the exponent.

Some other similar problems that I solved used Fermat's theorem (if N is prime, then $a^{N-1} \equiv$ 1 mod N), but 54 is not prime so I'm not sure if that is correct.

kd2amc
  • 11

2 Answers2

5

Hint: The problem is $(-1)^{1069}\mod 54$ and $1069$ is odd.

Wuestenfux
  • 20,964
1

$53 \equiv -1 \bmod 54$ thus we can have $53^{1069} \equiv (-1)^{1069} \equiv -1 \equiv 53 \bmod 54$.

Lily
  • 86