Number System

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Unit digit & cyclicity

⏱ 4 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

The unit digit of a power depends only on the unit digit of the base and on the exponent's remainder when divided by four. Unit digits of powers repeat in short cycles, so huge expressions collapse to one digit in seconds.

Overview

The unit digit is the last digit of a number, the ones place. Exams ask for the unit digit of monsters like $7^{105} \times 3^{58}$ , numbers far too big to compute. The rescue: the unit digit of a product depends only on the unit digits of the factors. $1237 \times 48$ ends the same way as $7 \times 8 = 56$ , that is, in $6$ .

Power cycles

Write out powers of $7$ : the unit digits run $7, 9, 3, 1$ , then repeat. This repeat length is the cyclicity.

Base digit	Cycle	Length
$0, 1, 5, 6$	never changes	$1$
$4$	$4, 6$	$2$
$9$	$9, 1$	$2$
$2$	$2, 4, 8, 6$	$4$
$3$	$3, 9, 7, 1$	$4$
$7$	$7, 9, 3, 1$	$4$
$8$	$8, 4, 2, 6$	$4$

Rule: Divide the exponent by $4$ . A remainder of $1$ , $2$ or $3$ means use that power; a remainder of $0$ means use the fourth power. For $4$ and $9$ , only odd or even matters.

The method step by step

Take $8^{67}$ . Keep the base digit $8$ . Divide $67$ by $4$ : remainder $3$ . The third member of the cycle of $8$ is $2$ , so the unit digit is $2$ . Only the last two digits of a big exponent matter, because $100$ divides by $4$ .

Products, sums and differences

Find each unit digit first, then combine. For sums, add the digits and keep the last one. For differences, subtract; if the result is negative, add $10$ . For $7^{105} - 3^{58}$ : the digits are $7$ and $9$ , and $7 - 9 = -2$ , so the answer is $10 - 2 = 8$ .

Watch: The digit-borrow trick for a difference assumes the first number really is the larger one. The question usually ensures this.

Free wins without cycles

Any even number times any number ending in $5$ ends in $0$ .
An odd number times a number ending in $5$ ends in $5$ .
Every factorial from $5!$ on ends in $0$ . So $1! + 2! + \cdots + 50!$ ends in $3$ , from $1 + 2 + 6 + 24 = 33$ .

Power towers

For $7^{21^{20}}$ you need the exponent $21^{20}$ by $4$ . Since $21$ leaves $1$ with $4$ , so does $21^{20}$ , and the unit digit is that of $7^1 = 7$ . For $3^{4^5}$ , the exponent $4^5$ is a multiple of $4$ , so use $3^4$ , giving $1$ .

Tip: For a tower, reduce the inner exponent first, then apply the outer cycle.

Last two digits

For a base ending in $1$ , like $71$ or $41$ : the last digit stays $1$ . The tens digit is the last digit of (tens digit of base times the exponent). For $71^{36}$ : $7 \times 36 = 252$ , keep $2$ . So the ending is $21$ . For $41^{27}$ : $4 \times 27 = 108$ , keep $8$ , ending $81$ .

Example: $6^{25} + 9^{31}$ ends in $6 + 9 = 15$ , so the unit digit is $5$ .

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Unit digit of a power or a product of powers

How to spot it:

The question asks for the unit digit of one big power or of a product of two or three powers.

\text{unit}(a^n) = \text{unit}\left(a^{\,n \bmod 4}\right)

Method

Keep only the unit digit of each base.
Reduce each exponent mod $4$ ; for base digits $4$ and $9$ , odd or even is enough.
Read each unit digit from its cycle.
Multiply the digits and keep the last one.

Why it works:

Unit digits of powers repeat every four steps or fewer, so only the exponent mod four matters.

Try this

Find the unit digit of 4 to the power 63 x 9 to the power 72 x 8 to the power 41.

Show solution

$4^{63}$ : odd power of $4$ gives $4$ . $9^{72}$ : even power of $9$ gives $1$ .
$8^{41}$ : $41 \bmod 4 = 1$ , cycle of $8$ starts at $8$ .
$4 \times 1 \times 8 = 32$ : unit digit $2$ .

Answer

Type 2common2 practice Q

Unit digit of a sum or difference of powers

How to spot it:

Powers joined by plus or minus, like 7 to the 105 minus 3 to the 58.

Method

Find the unit digit of each power separately.
Add or subtract those digits.
Keep the last digit; if a difference is negative, add $10$ .

Why it works:

Carrying and borrowing never change the ones place beyond what the unit digits show.

Try this

Find the unit digit of 7 to the power 105 minus 3 to the power 58.

Show solution

$7^{105}$ : $105 \bmod 4 = 1$ , digit $7$ .
$3^{58}$ : $58 \bmod 4 = 2$ , digit $9$ .
$7 - 9 = -2$ , add $10$ : digit $8$ .

Answer

Type 3common3 practice Q

Unit digit of long products and factorial sums

How to spot it:

A product of many plain numbers, a product like 1 x 3 x 5 x ..., or a factorial sum 1! + 2! + ... + n!.

Method

Scan for an even factor together with a factor ending in $5$ : answer $0$ at once.
For factorial sums, drop every term from $5!$ on.
Otherwise multiply unit digits step by step, keeping the last digit.

Why it works:

Two times five makes a ten, and a trailing zero survives every further multiplication.

Try this

Find the unit digit of 2 to the power 31 x 5 to the power 17 x 3 to the power 9.

Show solution

$2^{31}$ is even, and $5^{17}$ ends in $5$ .
Their product ends in $0$ .
Whatever multiplies it, the unit digit stays $0$ .

Answer

Type 4occasional2 practice Q

Power towers

How to spot it:

The exponent is itself a power, as in 7 to the power 21 to the power 20.

a^{b^c}: \text{reduce } b^c \bmod 4 \text{ first}

Method

Reduce the inner power modulo $4$ .
Use that remainder on the cycle of the base digit.
If the inner base is even, its power is usually a multiple of $4$ : use the fourth power.

Why it works:

Only the exponent's position in the four-cycle matters, whatever its size.

Try this

Find the unit digit of 7 to the power 21 to the power 20.

Show solution

$21 \equiv 1 \pmod 4$ , so $21^{20} \equiv 1 \pmod 4$ .
Use $7^1$ .
Unit digit $7$ .

Answer

Type 5occasional2 practice Q

Last two digits of a power

How to spot it:

The question asks for the last two digits of a power whose base ends in 1, such as 71 or 41.

(10a + 1)^n \text{ ends in } \left[(a \cdot n) \bmod 10\right] 1

Method

Confirm the base ends in $1$ ; the last digit of the answer is $1$ .
Multiply the base's tens digit by the exponent.
Keep the last digit of that product as the tens digit.

Why it works:

Expanding the power leaves the tens digit driven only by a times n, everything higher vanishes mod 100.

Try this

Find the last two digits of 71 to the power 36.

Show solution

Tens digit of base: $7$ ; exponent: $36$ .
$7 \times 36 = 252$ ; keep the last digit $2$ .
Last two digits: $21$ .

Answer

Formula sheet

Cyclicity rule

\text{unit}(a^n) = \text{unit}(a^r), \quad r = n \bmod 4 \ (r = 0 \Rightarrow r = 4)

Cycles of two, three, seven, eight

2: 2,4,8,6 \quad 3: 3,9,7,1 \quad 7: 7,9,3,1 \quad 8: 8,4,2,6

Factorials

n! \equiv 0 \pmod{10} \text{ for } n \ge 5

Last two digits, base ending in one

(10a + 1)^n \text{ ends in } \left[(a \cdot n) \bmod 10\right] 1

Shortcuts that save time

⚡ Last two digits of the exponent

For division by four, only the last two digits of the exponent matter, because 100 is a multiple of 4.

Example

Find the unit digit of 1357 to the power 2463.

Show solution

Base ends in $7$ ; only $63$ of the exponent matters.
$63 \bmod 4 = 3$ , so use $7^3 = 343$ .
Unit digit $3$ .

Answer

⚡ Even times five ends in zero

If a product contains an even factor and a factor ending in 5, the unit digit is 0 without any cycle work.

Example

Find the unit digit of 2 to the power 31 x 5 to the power 17 x 3 to the power 9.

Show solution

$2^{31}$ is even; $5^{17}$ ends in $5$ .
Even times a number ending in $5$ ends in $0$ .
The whole product ends in $0$ .

Answer

⚡ Factorial sums stop at four terms

From 5! on, every factorial ends in 0, so a factorial sum's unit digit comes from the first four terms alone.

Example

Find the unit digit of 1! + 2! + 3! + ... + 50!.

Show solution

Terms from $5!$ on end in $0$ .
$1 + 2 + 6 + 24 = 33$ .
Unit digit $3$ .

Answer

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Treating an exponent remainder of 0 as power 0.

Remainder 0 means use the fourth power of the cycle, not power 1.

Mistake 02

Reducing the exponent mod 4 using only its last digit.

Use the last two digits of the exponent, since 100 is the multiple of 4.

Mistake 03

Forgetting to borrow ten in a difference of unit digits.

7 minus 9 is negative: add 10 and answer 8.

Mistake 04

Applying the cycle table to the whole base number.

Only the last digit of the base enters the table.

Quick revision

Read this the night before the exam.

Unit digit needs only the base's last digit and $n \bmod 4$ .
Remainder $0$ means use the fourth power.
$0, 1, 5, 6$ never change; $4$ and $9$ have cycle $2$ .
Even $\times$ ending-5 gives $0$ ; odd $\times$ ending-5 gives $5$ .
$n!$ for $n \ge 5$ ends in $0$ .
Negative digit difference: add $10$ .
Base ending in $1$ : tens digit $= ($ base tens $\times$ exponent $) \bmod 10$ .

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 9 min · wrong answers go to your mistake notebook automatically.

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