Divisibility rules are quick mental tests that tell you whether one whole number divides exactly into another — without doing long division. Knowing the rules for 2, 3, 4, 5, 6, 8, 9, and 10 saves time in exams and is an essential KS3 number skill.
What does "divisible" mean?
A whole number is divisible by another if it divides in exactly, leaving no remainder. For example, 18 is divisible by 3 (18 ÷ 3 = 6 exactly), but 19 is not divisible by 3 (19 ÷ 3 = 6 remainder 1).
Divisibility is the same as asking whether the second number is a factor of the first. So knowing divisibility rules is a shortcut to spotting factors without performing the full calculation.
What are all the divisibility rules?
Here is a summary of the most useful divisibility rules at KS3.
| Divisor | Rule | Example |
|---|---|---|
| 2 | Last digit is 0, 2, 4, 6, or 8 (even) | 346 → last digit 6, even → divisible |
| 3 | Sum of all digits is divisible by 3 | 471 → 4+7+1 = 12, 12÷3 = 4 → divisible |
| 4 | Last two digits form a number divisible by 4 | 1,328 → 28 ÷ 4 = 7 → divisible |
| 5 | Last digit is 0 or 5 | 1,745 → last digit 5 → divisible |
| 6 | Divisible by BOTH 2 and 3 | 624 → even ✓, 6+2+4=12, 12÷3=4 ✓ → divisible |
| 8 | Last three digits form a number divisible by 8 | 5,624 → 624 ÷ 8 = 78 → divisible |
| 9 | Sum of all digits is divisible by 9 | 2,862 → 2+8+6+2 = 18, 18÷9 = 2 → divisible |
| 10 | Last digit is 0 | 4,730 → last digit 0 → divisible |
How do the digit-sum rules for 3 and 9 work?
The rules for 3 and 9 depend on the sum of digits, not just the last digit.
Divisibility by 3: Add all the digits together. If the result is divisible by 3, so is the original number.
Worked example: Is 5,814 divisible by 3? 5 + 8 + 1 + 4 = 18. Is 18 divisible by 3? Yes (18 ÷ 3 = 6). So 5,814 is divisible by 3.
If the digit sum is still large, add its digits again until you reach a single digit — this is called the digital root. A digital root of 3, 6, or 9 means divisible by 3; a digital root of exactly 9 means divisible by 9.
Worked example: Is 9,351 divisible by 9? 9 + 3 + 5 + 1 = 18. 1 + 8 = 9. Digital root is 9, so 9,351 is divisible by 9. Check: 9,351 ÷ 9 = 1,039. ✓
How do the last-two-digits and last-three-digits rules work?
For divisibility by 4, only the last two digits matter. This is because 100 is exactly divisible by 4, so any multiple of 100 is automatically divisible by 4. What is left to check is the remainder — the last two digits.
Worked example: Is 2,736 divisible by 4? Last two digits: 36. Is 36 ÷ 4 exact? 36 ÷ 4 = 9. Yes — so 2,736 is divisible by 4.
The same logic applies to divisibility by 8: 1,000 is exactly divisible by 8, so check only the last three digits.
Worked example: Is 7,512 divisible by 8? Last three digits: 512. 512 ÷ 8 = 64. Yes — so 7,512 is divisible by 8.
How do you test divisibility by 6?
Six = 2 × 3, and 2 and 3 share no common factors, so a number is divisible by 6 if and only if it is divisible by both 2 and 3.
Worked example: Is 4,326 divisible by 6?
- Even? Last digit is 6 — yes. ✓
- Digit sum? 4 + 3 + 2 + 6 = 15. 15 ÷ 3 = 5 — yes. ✓
Both tests pass, so 4,326 is divisible by 6.
Counter-example: Is 4,332 divisible by 6? Last digit 2 — even ✓. Digit sum 4+3+3+2 = 12, divisible by 3 ✓. Both pass — yes it is divisible by 6.
What about 3,328? Last digit 8 — even ✓. Digit sum 3+3+2+8 = 16. 16 ÷ 3 = 5 remainder 1 — not divisible by 3. So 3,328 is not divisible by 6, even though it is divisible by 2.
When are divisibility rules useful in KS3 maths?
Divisibility rules are most useful when:
- Finding factors of a number without trial division
- Simplifying fractions — spot common factors before cancelling
- Prime factor decomposition — decide which prime to divide by first
- HCF and LCM problems — quickly identify shared factors
- Checking calculator-free exam answers — estimate whether a result makes sense
Worked example: Simplify the fraction 324/756.
- Is 324 divisible by 4? Last two digits 24 ÷ 4 = 6. Yes.
- Is 756 divisible by 4? Last two digits 56 ÷ 4 = 14. Yes.
- Divide both by 4: 81/189.
- Digit sum of 81: 8+1 = 9. Divisible by 9. ✓
- Digit sum of 189: 1+8+9 = 18. Divisible by 9. ✓
- Divide both by 9: 9/21. Then divide by 3: 3/7. ✓
Frequently asked questions
Is there a divisibility rule for 7?
Yes, but it is complicated enough that most students find it quicker to just divide. One method: double the last digit, subtract it from the remaining number, and repeat until you can judge easily. For example, to test 833: double 3 → 6, subtract from 83 → 77. Is 77 divisible by 7? Yes (77 ÷ 7 = 11). So 833 is divisible by 7. In practice, the rule for 7 is rarely tested at KS3 — knowing 2, 3, 4, 5, 6, 8, 9, 10 is enough.
Why does the digit-sum rule work for 3 and 9?
Any digit d in the tens position represents 10d = (9+1)d = 9d + d. The 9d part is always divisible by 9 (and by 3). So the divisibility of the whole number by 3 or 9 depends only on what is left after stripping out multiples of 9 — which is exactly the sum of the digits. The same reasoning extends to hundreds (99d + d) and larger place values.
Can I use these rules on decimals?
Divisibility rules apply to whole numbers only. If you need to check whether a decimal divides into something exactly, convert to a fraction or use long division.
What if the digit-sum test gives a two-digit answer?
Apply the digit-sum test again. For example, to check 9,999 for divisibility by 9: digit sum = 36. Then 3 + 6 = 9. Divisible by 9. You can keep applying the rule until you reach a single digit — this process always terminates.
For Socratic KS3 number practice with Professor Pi, see aitutors.me.