17. $[\mathbf{1 0}]$ Determine the last two digits of $17^{17}$, written in base 10 .
Answer: $\quad 77$ We are asked to find the remainder when $17^{17}$ is divided by 100 . Write the power as $(7+10)^{17}$ and expand with the binomial theorem:
$$
(7+10)^{17}=7^{17}+17 \cdot 7^{16} \cdot 10+\ldots
$$

We can ignore terms with more than one factor of 10 because these terms are divisible by 100 , so adding them does not change the last two digits. Now, $7^{4}=2401$, which has remainder 1 mod 100, so $7^{17}$ has last two digits 07 and $7^{16} \cdot 10$ has last two digits 70 . We add these together.