1. What is the last digit of $17^{103}+5$ ?
Solution: Let $\langle N\rangle$ be the last digit of $N$. $\left\langle 17^{2}\right\rangle=9,\left\langle 17^{3}\right\rangle=3,\left\langle 17^{4}\right\rangle=1$, and $\left\langle 17^{5}\right\rangle=7$. Since this pattern keeps on repeating itself, $\left\langle 17^{4 N}\right\rangle=1$ for any integer $N$. Thus $\left\langle 17^{2 \cdot 25}\right\rangle=$ $\left\langle 17^{100}\right\rangle=1$, so $\left\langle 17^{103}\right\rangle=3$, and $\left\langle 17^{103}+5\right\rangle=\langle 3+5\rangle=8$.