2018 AMC 10A Problem 19

There are $5 \cdot 20 = 100$ equally likely pairs $(m, n)$. The units digit of $m^n$ depends only on the units digit of $m$, so handle each $m$ separately. The $20$ values of $n$ form a complete block of $20$ consecutive integers, containing $10$ even numbers and $5$ multiples of $4$ ($2000, 2004, 2008, 2012, 2016$).

• $m = 11$: units digit $1$ for every $n$. $20$ favorable.
• $m = 13$: units digits of $3^n$ cycle $3, 9, 7, 1$; we need $4 \mid n$. $5$ favorable.
• $m = 15$: $5^n$ always ends in $5$. $0$ favorable.
• $m = 17$: $7^n$ cycles $7, 9, 3, 1$; need $4 \mid n$. $5$ favorable.
• $m = 19$: $9^n$ cycles $9, 1$; need $n$ even. $10$ favorable.

Total favorable pairs: $20 + 5 + 0 + 5 + 10 = 40$, so the probability is $\dfrac{40}{100} = \dfrac{2}{5}$.

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

Units digits of powers are periodic with period dividing $4$, so the question for each base is "how often does $n$ land on the right residue mod 4 (or mod 2)?" Because $1999$ to $2018$ is exactly $20 = 5 \cdot 4$ consecutive integers, each residue class mod $4$ appears exactly $5$ times, which makes the counting clean. Splitting by $m$ and adding is the natural casework; there is no shortcut across the different bases.

• Treating the five bases as behaving alike (all cycle length 4) and answering 1/4, or miscounting the multiples of 4 in 1999-2018.
• Averaging probabilities incorrectly, e.g. adding $1 + \tfrac14 + 0 + \tfrac14 + \tfrac12$ and forgetting to divide by $5$.