Technique

Unit digit of a power from its cycle

Arithmetic • Number System • Unit Digit

Statement

The unit digits of

a^{1}, a^{2}, a^{3}, \dots

repeat in a cycle of length

L

with

L \leq 4

; to find the unit digit of

a^{n}

, divide

n

L

and take the entry at position

r

when the remainder

r

is non-zero, or the last entry when

r = 0

How to use it

1Write out $a^{1}$ to $a^{4}$ and stop as soon as the first unit digit reappears; that fixes the cycle.
2The digits $0, 1, 5, 6$ have cycle length $1$ ; $4$ and $9$ have length $2$ ; $2, 3, 7, 8$ have length $4$ .
3Count the cycle from $a^{1}$ and not from $a^{0}$ , and send a remainder of $0$ to the last entry.

What is the unit digit of

3^{50}

Step-by-step

1
$3^{1}, 3^{2}, 3^{3}, 3^{4}$ end in $3, 9, 7, 1$
Why: The digit $1$ returns at the fourth power, so the cycle has length $4$ .
2
$50 = 4 \times 12 + 2$
Why: Only the remainder on division by the cycle length matters.
3
The second entry of the cycle is $9$ .

Answer

9

Watch out

A remainder of

0

points at the last entry of the cycle, not the first: that is where the off-by-one lives. Only the unit digit of the base matters, so

1 3^{n}

and

3^{n}

end alike.

Also known as

cyclicity, last digit of a power, units digit pattern, power cycle, unit digit