A number is divisible by another when the division leaves no remainder. The rules for 2 through 11 decide it by looking at the last digits or at the sum of the digits, with no division. The number 2340 works well as an example because it passes almost every rule and fails only three: it is divisible by 2, 3, 4, 5, 6, 9 and 10, but not by 7, 8 or 11.
| Divisor | It holds if | Example |
|---|---|---|
| 2 | The last digit is even | 538 ends in 8, and 538 = 2 × 269 |
| 3 | The sum of the digits is a multiple of 3 | 4 + 7 + 1 = 12, and 471 = 3 × 157 |
| 4 | The last two digits form a multiple of 4 | 24 = 4 × 6, and 124 = 4 × 31 |
| 5 | It ends in 0 or 5 | 835 = 5 × 167 |
| 6 | It passes the rules for 2 and for 3 | 72 is even and 7 + 2 = 9, and 72 = 6 × 12 |
| 7 | Drop the last digit and subtract twice that digit from the rest to get a multiple of 7 | 34 − 2 × 3 = 28, and 343 = 7 × 49 |
| 8 | The last three digits form a multiple of 8 | 728 = 8 × 91, and 2728 = 8 × 341 |
| 9 | The sum of the digits is a multiple of 9 | 3 + 4 + 5 + 6 = 18, and 3456 = 9 × 384 |
| 10 | It ends in 0 | 580 = 10 × 58 |
| 11 | The digits with alternating signs add up to 0 or a multiple of 11 | 5 − 3 + 1 − 3 = 0, and 3135 = 11 × 285 |
The rules for 2, 5 and 10 look only at the last digit, the rule for 4 looks at two digits, and the rule for 8 looks at three. The reason is that 10 is a multiple of 2 and of 5, 100 is a multiple of 4, and 1000 is a multiple of 8, so everything in front of those digits is already divisible. The rules for 3, 9 and 11 need every digit, and the next section shows why.
Why the rules for 3, 9 and 11 work
Powers of 10 leave remainder 1 when divided by 9, because 10 = 9 + 1, 100 = 99 + 1 and 1000 = 999 + 1. That is why a number and the sum of its digits leave the same remainder when divided by 9. In 2340 = 2 × 999 + 3 × 99 + 4 × 9 + 9, the first three terms are multiples of 9, so only the digit sum is left to check. Since 999, 99 and 9 are also multiples of 3, the same argument works for the rule for 3.
If the digit sum is still large, add its digits again. In 987,654, the digits add up to 39, and the digits of 39 add up to 12. That 12 is a multiple of 3 but not of 9, so 987,654 is divisible by 3 and not by 9. Every multiple of 9 is a multiple of 3, but not the other way around.
The rule for 11 follows the same idea with a different remainder. Since 10 = 11 − 1, the number 10 leaves remainder −1, 100 leaves remainder 1, 1000 leaves remainder −1, and so on, alternating. Adding the digits with alternating signs therefore gives the same remainder as the whole number. In 2728, 2 − 7 + 2 − 8 = −11, and 2728 = 11 × 248. In 1001 you have to alternate through the zeros too, 1 − 0 + 0 − 1 = 0, which confirms that 1001 = 11 × 91. The sign of the result does not matter, because −11 and 11 are both multiples of 11.
Worked example: 2340 against all ten rules
Apply the rules for 2 through 11 to the number 2340 and note which ones it is divisible by.
- Look at the end. It ends in 0, so it is divisible by 2, 5 and 10. The last two digits make 40, which is 4 × 10, so it is also divisible by 4.
- Add the digits. 2 + 3 + 4 + 0 = 9, a multiple of 3 and of 9. Since the number is also even, it passes the rule for 6.
- Check 8. The last three digits make 340, and 340 = 8 × 42 + 4. The remainder is 4, so it is not divisible by 8, even though it is divisible by 4.
- Check 7. Drop the final 0 and 234 is left. Drop the 4 and subtract twice that digit from 23, which leaves 15, not a multiple of 7.
- Check 11. Alternate the signs of the digits to get 2 − 3 + 4 − 0 = 3. That is neither 0 nor a multiple of 11.
Result. 2340 is divisible by 2, 3, 4, 5, 6, 9 and 10, and it is not divisible by 7, 8 or 11.
Check. 2340 = 9 × 260, and 2340 ÷ 8 gives 292.5, which is not a whole number. Dividing by 7 leaves remainder 2, because 7 × 334 = 2338.
Composite divisors: 6, 12 and 15
For a composite divisor, check two rules for coprime divisors, meaning divisors with no common factor, whose product is the divisor. 6 is 2 × 3, 12 is 3 × 4 and 15 is 3 × 5. 2340 ends in 0 and its digits add up to 9, so it passes the rules for 3 and for 5, and that is why 2340 = 15 × 156.
The coprime condition keeps you from counting the same factor twice. You do not check 12 as 2 × 6, because 2 and 6 share the factor 2. You can also see it with 4 and 3, which give 2340 = 12 × 195.
Two other mix-ups give a false result. For the rule for 4, the last digit is not enough, because 1114 ends in 4 and still leaves remainder 2, since 1114 = 4 × 278 + 2. And the digit sum only tells you about 3 and 9, because 16 is a multiple of 4 and of 8 while 1 + 6 = 7.
Using the rules to simplify a fraction
When you simplify, the rules help you find a common divisor without guessing. In 45/60, both numbers end in 0 or 5, so they can be divided by 5, which leaves 9/12. The digits of 9 and of 12 add up to 9 and 3, multiples of 3, so they can be divided by 3, which leaves 3/4. You divided twice, by 5 and by 3, which is the same as dividing by 15 in one step. That 15 is the greatest common factor (GCF, also called the GCD) of 45 and 60, and the resulting fraction is equivalent to the original, as the guide to equivalent fractions explains.
With large terms, the rules only go up to 11. In 1001/1430, the rule for 11 divides both terms and leaves 91/130. The common factor 13 that remains is not detected by any rule from 2 to 11, which is why it is hard to spot at a glance. The greatest common factor of 1001 and 1430 is 143, and with it you get to 7/10 in one step. If you would rather see how to find it without rules, the guide to the Euclidean algorithm explains it.
To reduce any fraction instantly, use the simplify fractions calculator. The prime factorization calculator breaks a number into primes, and the factors calculator lists all of a number’s positive divisors, so it works for checking the result of a rule. For 2340, the list includes 2, 3, 4, 5, 6, 9 and 10, and it does not include 7, 8 or 11. To choose between GCF and LCM for a task, see GCD vs. LCM.
Practice
- Is 1248 divisible by 8?
- Is 1248 divisible by 9?
- Is 4059 divisible by 11?
- Is 5316 divisible by 12?
- Simplify 84/126 by dividing by the divisors the rules detect.
Show the answers
- Yes. The last three digits make 248 = 8 × 31, and 1248 = 8 × 156.
- No. 1 + 2 + 4 + 8 = 15, which is not a multiple of 9.
- Yes. 4 − 0 + 5 − 9 = 0, and 4059 = 11 × 369.
- Yes. 5 + 3 + 1 + 6 = 15 passes the rule for 3, and 16 = 4 × 4 passes the rule for 4, and 5316 = 12 × 443.
- Both are even, which leaves 42/63. The digits of 42 and of 63 add up to 6 and 9, so both can be divided by 3, which leaves 14/21. Both are multiples of 7, which leaves 2/3.