Question 1: What are the last two digits of $17^{198}$?
My Attempt: The pdf hinted to reduce it via mod $100$. So my work is as follows:
Since $17^2\equiv289\equiv-11\mod100$, we have$$17^{198}\equiv (-11)^{99}\equiv-11^{99}\mod100$$
However, I'm not sure how to compute $-11^{99}$ without too much of a hassle.
Questions:
- How do you complete this problem?
- Why use mod $100$? Why not mod $50$ or some other number?
Best Answer
An alternative approach, exploiting the Chinese remainder theorem and the fact that $2$ is a generator in $\mathbb{Z}/(25\mathbb{Z})^*$:
$$ 17^{198}\equiv (-8)^{198} \equiv 2^{3\cdot 198} \equiv 2^{14} \equiv 16\cdot 1024\equiv -16\equiv 9\pmod{25} $$ $$ 17^{198}\equiv 1\pmod{4} $$ together imply $17^{198}\equiv \color{red}{9}\pmod{100}$.