Math Core

The units digit of a number is its remainder when divided by $10$ . That makes units-digit questions a special kind of remainder problem, and one of the most common number theory questions on MATHCOUNTS and the AMC 8. The key fact: only units digits affect units digits.

Only the last digit matters

When you add or multiply, the units digit of the answer depends only on the units digits of the inputs. For $347 \times 86$ , the units digit is the units digit of $7 \times 6 = 42$ , which is $2$ . The tens, hundreds and higher digits only affect the higher digits of the result.

So in any units-digit problem, you can replace every number by its last digit right away: $2026^{5}$ has the same units digit as $6^5$ .

Powers cycle

Here are the units digits of the first few powers of each digit.

Last digit	Units digits of powers	Cycle length
$0, 1, 5, 6$	always $0$ , $1$ , $5$ , $6$	$1$
$4$	$4, 6, 4, 6, \dots$	$2$
$9$	$9, 1, 9, 1, \dots$	$2$
$2$	$2, 4, 8, 6, \dots$	$4$
$3$	$3, 9, 7, 1, \dots$	$4$
$7$	$7, 9, 3, 1, \dots$	$4$
$8$	$8, 4, 2, 6, \dots$	$4$

Every cycle length divides $4$ . So for any last digit, the pattern repeats every $4$ powers.

Units digit of a power

To find the units digit of $a^n$ (for $n \ge 1$ ), find the remainder when $n$ is divided by $4$ .

Remainder $1, 2, 3$ : use the $1$ st, $2$ nd or $3$ rd entry of the cycle for the last digit of $a$ .
Remainder $0$ : use the $4$ th entry (the last one in the cycle).

Common mistake

Remainder $0$ means the end of the cycle, not "the units digit is $1$ " or "the $0$ th power." For $2^{20}$ : $20$ leaves remainder $0$ , so use the $4$ th entry of $2, 4, 8, 6$ , which is $6$ .

Worked example: A single power

What is the units digit of $3^{45}$ ?

$45 = 4 \cdot 11 + 1$ , remainder $1$ . The cycle for $3$ is $3, 9, 7, 1$ , so use the $1$ st entry: $3$ .

Worked example: A sum of powers

What is the units digit of $2^{20} + 3^{30} + 7^{40}$ ?

$2^{20}$ : $20$ leaves remainder $0$ , so the $4$ th entry of $2, 4, 8, 6$ : $6$ .
$3^{30}$ : $30$ leaves remainder $2$ , so the $2$ nd entry of $3, 9, 7, 1$ : $9$ .
$7^{40}$ : $40$ leaves remainder $0$ , so the $4$ th entry of $7, 9, 3, 1$ : $1$ .

Add the units digits: $6 + 9 + 1 = 16$ . The units digit is $6$ .

Worked example: A tower of exponents

What is the units digit of $3^{3^3}$ ?

A tower is read from the top down: $3^{3^3} = 3^{27}$ . Since $27 = 4 \cdot 6 + 3$ , use the $3$ rd entry of $3, 9, 7, 1$ : the units digit is $7$ .

For a much taller tower, you only need the exponent's remainder when divided by $4$ . Find that with remainder rules.

Tip

A product that includes a factor ending in $5$ and an even factor ends in $0$ . A product of only odd numbers that includes one ending in $5$ ends in $5$ . Check for these before doing any work.