modulus magic............

Number Theory Level 1

find the value of

                    (996)^996  mod 997

note : here "mod" denotes the "%" or "modulus operator".....

1 solution

Prasun Biswas
Nov 22, 2014

Here, first of all, we can see that $997$ is a prime. Now,

$(996)^{996} = (996)^{997-1}$

Since $996$ is not divisible by $997$ , using Fermat's Little Theorem , with $a=996,p=997$ we have,

$a^{p-1} \equiv 1 \quad (\mod p)$

So, the required remainder $=\boxed{1}$

there was no need of any theorem because we see that $996\equiv(-1)\pmod{997}\\ \Rightarrow996^{996}\equiv1\pmod{997}\ because\ 996\ is\ even.$

Adarsh Kumar - 6 years, 6 months ago