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2018 AMC 10A Problem 19

2018 AMC 10A Problem 19AMC 10 Step by Step

Name: Date: October 28, 2026

Modular Arithmetic · difficulty 3 of 5 · about 4 min · position P16-20

A number $m$ is randomly selected from the set $\{11,13,15,17,19\}$ , and a number $n$ is randomly selected from $\{1999,2000,2001,\ldots,2018\}$ . What is the probability that $m^n$ has a units digit of $1$ ?

A) $\frac{1}{5}$
B) $\frac{1}{4}$
C) $\frac{3}{10}$
D) $\frac{7}{20}$
E) $\frac{2}{5}$

Answer

Key insight

Only the last digit of m matters: 1 always works, 5 never, 9 needs an even exponent, 3 and 7 need a multiple of 4.

Solution

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.

List the last digits of small powers.

last digit of $m$	$m^1$	$m^2$	$m^3$	$m^4$	$m^5$
$1$	$1$	$1$	$1$	$1$	$1$
$3$	$3$	$9$	$7$	$1$	$3$
$5$	$5$	$5$	$5$	$5$	$5$
$7$	$7$	$9$	$3$	$1$	$7$
$9$	$9$	$1$	$9$	$1$	$9$

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

Why this works

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.

The trap

Assuming every base repeats every 4 steps the same way and answering 1/4, or miscounting how many multiples of 4 lie between 1999 and 2018 (there are 5).

Concise solution

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

Review in the AoPS books

Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)

Techniques

Split into exhaustive cases and handle each · Compute small cases, spot the pattern, generalize

last digit of $m$	$m^1$	$m^2$	$m^3$	$m^4$	$m^5$
$1$	$1$	$1$	$1$	$1$	$1$
$3$	$3$	$9$	$7$	$1$	$3$
$5$	$5$	$5$	$5$	$5$	$5$
$7$	$7$	$9$	$3$	$1$	$7$
$9$	$9$	$1$	$9$	$1$	$9$

last digit of $m$	$m^1$	$m^2$	$m^3$	$m^4$	$m^5$
$1$	$1$	$1$	$1$	$1$	$1$
$3$	$3$	$9$	$7$	$1$	$3$
$5$	$5$	$5$	$5$	$5$	$5$
$7$	$7$	$9$	$3$	$1$	$7$
$9$	$9$	$1$	$9$	$1$	$9$

last digit of $m$	$m^1$	$m^2$	$m^3$	$m^4$	$m^5$
$1$	$1$	$1$	$1$	$1$	$1$
$3$	$3$	$9$	$7$	$1$	$3$
$5$	$5$	$5$	$5$	$5$	$5$
$7$	$7$	$9$	$3$	$1$	$7$
$9$	$9$	$1$	$9$	$1$	$9$