This is what I have so far:
Since 101 is a prime and does not divide 7, we can apply Fermat's Little Theorem to see that $$7^{100} \equiv 1 \ (mod \ 101)$$
We can then reduce $7^{2002}$ to $7^{2} (7^{100})^{20} \equiv 7^{2}(1)^{20} \ (mod \ 101)$ which is where I'm stuck at.
$7^2=49 \equiv 150 \ (mod \ 101)$. How do I reduce $7^2$ in a way that is constructive towards my solution since $(mod \ 101)$ is such a large modulus to operate in?
[Math] Find the remainder of $7^{2002}$ divided by 101.
elementary-number-theory
Best Answer
You had the solution and broke it :)
At
$$7^{2002} \equiv 49\pmod {101} \,,$$
you are done, the remainder must be $49$. Indeed, if you denote the remainder by $r$ then $0 \leq r \leq 100$ and
$$ r \equiv 49 \pmod{101} \,.$$
This means that $101|r-49$, and since $-49 \leq r-49 < 52$ you get $r-49=0$.