FactorsDivisibility rules from 2 to 12, with reasons
A divisibility rule tells you whether one number divides another without doing the division. The rules save time in finding factors, simplifying fractions and checking answers. Each rule below has an example and a short reason, because a rule you understand is a rule you do not forget.
| By | The number divides if | Example |
|---|---|---|
| 2 | its last digit is 0, 2, 4, 6 or 8 | 3,58: yes |
| 3 | the sum of its digits is a multiple of 3 | 4,812: 4 + 8 + 1 + 2 = 15, yes |
| 4 | its last two digits form a multiple of 4 | 7,316: 16 ÷ 4 = 4, yes |
| 5 | its last digit is 0 or 5 | 2,345: yes |
| 6 | it is divisible by both 2 and 3 | 4,812: even, digits add to 15, yes |
| 7 | doubling the last digit and taking it from the rest gives a multiple of 7 | 343: 34 − 6 = 28, yes |
| 8 | its last three digits form a multiple of 8 | 5,120: 120 ÷ 8 = 15, yes |
| 9 | the sum of its digits is a multiple of 9 | 7,263: 7 + 2 + 6 + 3 = 18, yes |
| 10 | its last digit is 0 | 4,560: yes |
| 11 | the difference between the sums of alternate digits is 0 or a multiple of 11 | 9,174: (9 + 7) − (1 + 4) = 11, yes |
| 12 | it is divisible by both 3 and 4 | 7,316: digits add to 17, no |
Why the last-digit rules work (2, 5, 10)
Any number is a pile of tens plus its last digit: 358 = 35 tens + 8. Tens are always divisible by 2, 5 and 10. So only the last digit decides.
Why the rules for 4 and 8 work
A hundred is 4 × 25, so every pile of hundreds is divisible by 4 and only the last two digits matter. A thousand is 8 × 125, so for 8 only the last three digits matter.
Why the digit-sum rules work (3 and 9)
10 is 9 + 1, 100 is 99 + 1 and 1000 is 999 + 1. So 4,812 is 4 × 999 + 8 × 99 + 1 × 9, plus 4 + 8 + 1 + 2. The first part is always divisible by 9 and by 3. What is left over is the digit sum, and it alone decides.
Why the rule for 11 works
10 is one less than 11, 100 is one more than 99 (a multiple of 11), 1000 is one less than 1001 (also a multiple of 11), and so on in turn. So digits in alternate places count as plus and minus. If the two sums cancel out, or differ by a multiple of 11, the number divides by 11.
Rules made from two rules (6, 12 and others)
To test for 6, test for 2 and 3. To test for 12, test for 3 and 4. For 15, test 3 and 5. For 18, test 2 and 9. The two numbers you split into must be co-prime. That is why 12 uses 3 and 4, not 2 and 6: the number 18 is divisible by 2 and by 6 but not by 12.
The rule for 7
Double the last digit and subtract it from the number formed by the other digits. If the result is 0 or a multiple of 7, so is the number. For 1,274: 127 − 8 = 119. Again: 11 − 18 = −7. That is a multiple of 7, so 1,274 is divisible by 7. For small numbers, dividing by 7 directly is usually just as quick.
Using the rules together
Is 2,520 divisible by every number from 2 to 10? It ends in 0, so 2, 5 and 10 pass. Its digits add to 9, so 3 and 9 pass, and with 2 that gives 6. Its last two digits, 20, give 4. Its last three, 520, are 8 × 65. And 252 − 0 = 252 = 7 × 36. Yes: 2,520 is in fact the smallest number divisible by all of 1 to 10.
Every "factors of" page on this site, such as factors of 360, walks through these rules for that number.