101 is a prime number. What is the remainder when $3^{101}$ is divided by 101?

Fermat's little theorem says that for a prime $p$ and any whole number a, the number $a^p$ leaves the same remainder as a itself on division by $p$. With $p=101$ and $a=3$, the remainder is simply 3. Put another way, $3^{100}$ ≡ 1 (mod 101), so $3^{101}=3^{100}\times 3$ ≡ 3. The trap is 1: that is the remainder left by $3^{100}$, one power lower down.