πŸ”’
MathπŸ”¬ Ages 11-13Intermediate 10 min read

Divisibility Rules

Quick divisibility tests for 2, 3, 4, 5, 6, 8, 9, 10 and 11 β€” how to tell if a number divides exactly without doing the division, why each rule works, with worked examples and a quiz.

Key takeaways

  • Divisibility rules tell you if a number divides exactly, without doing the full division
  • A number is divisible by 3 (or 9) if its digit sum is divisible by 3 (or 9)
  • The last digit settles divisibility by 2, 5 and 10; the last digits handle 4 and 8
  • Combine rules: divisible by 6 means divisible by both 2 and 3

Checking without dividing

A divisibility rule is a quick test that tells you whether one number divides exactly into another β€” that is, with no remainder β€” without doing the whole division. These tests save time, help you simplify fractions, find Factors and Multiples, and spot Prime Numbers.

A number is divisible by another when dividing leaves a remainder of zero. For example, 24 is divisible by 6 because 24 Γ· 6 = 4 exactly, but 25 is not. Let's collect the most useful rules and, crucially, understand why each one works.

The last-digit rules: 2, 5 and 10

These are the easiest because only the final digit matters.

  • Divisible by 2 if the last digit is even (0, 2, 4, 6 or 8).
  • Divisible by 5 if the last digit is 0 or 5.
  • Divisible by 10 if the last digit is 0.

Why? Every place value above the units β€” tens, hundreds, thousands β€” is already a multiple of 2, 5 and 10. So the only thing that can spoil divisibility is the units digit. If the units digit passes, the whole number passes.

The digit-sum rules: 3 and 9

  • Divisible by 3 if the sum of the digits is divisible by 3.
  • Divisible by 9 if the sum of the digits is divisible by 9.

Example β€” is 5,742 divisible by 3? Add the digits: 5 + 7 + 4 + 2 = 18. Since 18 Γ· 3 = 6, yes, 5,742 is divisible by 3. (And because 18 is also divisible by 9, 5,742 is divisible by 9 too.)

Why does this work? Think about place value. 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, and so on. The 9, 99 and 999 parts are all multiples of 9 (and of 3), so they never affect the remainder. What is left over from each digit is just the digit itself. So the remainder of the whole number depends only on the sum of the digits. That is the elegant reason behind the rule.

The last-two and last-three rules: 4 and 8

  • Divisible by 4 if the number formed by the last two digits is divisible by 4.
  • Divisible by 8 if the number formed by the last three digits is divisible by 8.

Example β€” is 7,316 divisible by 4? Look only at the last two digits: 16. Since 16 Γ· 4 = 4, yes.

Why? Because 100 is divisible by 4, every hundred (and thousand, etc.) is already a multiple of 4. So only the last two digits can decide the question. Similarly, 1000 is divisible by 8, so only the last three digits matter for 8.

The combination rules: 6

  • Divisible by 6 if the number is divisible by both 2 and 3.

Why? Because 6 = 2 Γ— 3, and 2 and 3 share no common factor, a number divisible by both must be divisible by their product, 6. So you do not need a special rule for 6 β€” just use the rules for 2 and 3 together.

Example β€” is 534 divisible by 6? It is even (passes 2), and 5 + 3 + 4 = 12 is divisible by 3 (passes 3). Both pass, so 534 is divisible by 6.

The alternating-sum rule: 11

This one is less common but clever. A number is divisible by 11 if the alternating sum of its digits (subtract, add, subtract, ...) is 0 or a multiple of 11.

Example β€” is 3,729 divisible by 11? Take the digits and alternate signs from the right: 9 βˆ’ 2 + 7 βˆ’ 3 = 11. Since 11 is a multiple of 11, yes β€” 3,729 Γ· 11 = 339.

A reference table

Divisible byTestExample (passes)
2Last digit is even158
3Digit sum divisible by 3213 (2+1+3=6)
4Last two digits divisible by 41,128 (28)
5Last digit 0 or 54,075
6Divisible by 2 and 3522
8Last three digits divisible by 85,016 (016)
9Digit sum divisible by 96,381 (6+3+8+1=18)
10Last digit 03,490
11Alternating digit sum is 0 or Γ· 112,728

Worked example

Is 8,532 divisible by 2, 3, 4, 6 and 9? Test each.

  1. By 2: last digit 2 is even β†’ yes.
  2. By 3: 8 + 5 + 3 + 2 = 18, divisible by 3 β†’ yes.
  3. By 4: last two digits 32, and 32 Γ· 4 = 8 β†’ yes.
  4. By 6: divisible by both 2 and 3 β†’ yes.
  5. By 9: digit sum 18 is divisible by 9 β†’ yes (8,532 Γ· 9 = 948).

So 8,532 passes every one of those tests. A single number can satisfy many rules at once.

A practice activity

Make a "divisibility detective" grid. Across the top write 2, 3, 4, 5, 6, 9, 10. Down the side write these numbers: 90, 144, 255, 612, 1,000. For each number, tick every column it is divisible by, using the rules above.

Then check yourself: 144 should tick 2, 3, 4, 6 and 9; 255 should tick only 3 and 5; 1,000 should tick 2, 4, 5 and 10.

Where this leads

Divisibility rules make it fast to break a number into its building blocks, which is the heart of finding factors, simplifying fractions, and identifying Prime Numbers. They reward you for understanding place value rather than just grinding through long division β€” a theme that runs right through higher maths.

Quick quiz

Test yourself and earn XP

Is 4,128 divisible by 3?

Which number is divisible by both 2 and 3 (and so by 6)?

How do you test divisibility by 4?

Is 90 divisible by 9?

A number ends in 5. Which rule does it definitely pass?

FAQ

They come from place value. For 2, 5 and 10 only the last digit matters because every higher place (10, 100, ...) is already a multiple of those numbers. For 3 and 9 the digit-sum rule works because 10, 100 and 1000 are all one more than a multiple of 9.

There is, but it is awkward: double the last digit, subtract it from the rest, and check if the result is divisible by 7. Most people find it quicker to just divide by 7 directly, which is why the 7 rule is rarely used.