2018 AMC 10A Problem 19

There are $5 \cdot 20 = 100$ equally likely pairs $(m, n)$. The last digit of $m^n$ depends only on the last digit of $m$, so look at each $m$ on its own.

1. List the last digits of small powers.

The patterns repeat, so:
- last digit $1$: always ends in $1$;
- last digit $3$ or $7$: ends in $1$ exactly when $n$ is a multiple of $4$;
- last digit $5$: never ends in $1$;
- last digit $9$: ends in $1$ exactly when $n$ is even.
2. Count the useful values of $n$. From $1999$ to $2018$ there are $20$ numbers. Of these, $10$ are even, and $5$ are multiples of $4$: $2000, 2004, 2008, 2012, 2016$.
3. Count favorable pairs for each $m$.
- $m = 11$: all $20$ values of $n$.
- $m = 13$: $5$ values.
- $m = 15$: $0$ values.
- $m = 17$: $5$ values.
- $m = 19$: $10$ values.
4. Add and divide. $20 + 5 + 0 + 5 + 10 = 40$ favorable pairs out of $100$, so the probability is $\dfrac{40}{100} = \dfrac25$.

The answer is $\boxed{\textbf{(E)}\ \frac{2}{5}}$.

Last digits of powers repeat in a short cycle (at most $4$ long), so whether $m^n$ ends in $1$ depends only on where $n$ falls in that cycle. The range $1999$ to $2018$ is exactly $20$ numbers, which is five full cycles of $4$, so the counts come out evenly. Different last digits have different cycles, so handle each $m$ separately and add.