2017 AMC 10B Problem 14

Only the remainder of $N$ mod $5$ matters. Compute fourth powers of each nonzero residue:
$$1^4 = 1,\quad 2^4 = 16 \equiv 1,\quad 3^4 = 81 \equiv 1,\quad 4^4 = 256 \equiv 1 \pmod 5.$$
So whenever $5 \nmid N$, $N^{16} = (N^4)^4 \equiv 1^4 = 1 \pmod 5$. If $5 \mid N$, then $N^{16} \equiv 0$.

Thus $N^{16}$ leaves remainder $1$ exactly when $N$ is not a multiple of $5$. Since $2020 = 5 \cdot 404$, the range $1$ to $2020$ contains $404$ multiples of $5$ and $2020 - 404 = 1616$ non-multiples. The probability is
$$\frac{1616}{2020} = \frac{4}{5}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{4}{5}}$.

This is Fermat's little theorem in disguise: for a prime $p$ and $p \nmid N$, $N^{p-1} \equiv 1 \pmod p$. Even without knowing the theorem, checking the four nonzero residues takes seconds. The exponent $16$ is a multiple of $4$, which is the only reason it looks scary. Because $2020$ is divisible by $5$, the residues are perfectly balanced and the probability is a clean $\frac45$.

• Testing only N = 1, 2 and concluding the remainder depends on N in a complicated way, or forgetting that multiples of 5 give remainder 0.
• Answering $1$ (choice E) after checking a few non-multiples of $5$ and never trying $N = 5$.