Divisibility rules are guidelines that allow you to determine if a number is divisible by another number without performing the division: you just need to look at the last digits or sum all the digits of the number. This page contains a table of divisibility rules for 2, 3, 4, 5, 6, 8, 9, 10, 11, and 25, a clear explanation of the rules for 3 and 9 using the sum of digits, examples with solutions, common mistakes, and an interactive trainer where you can test yourself immediately.
What is a divisibility rule?
A divisibility rule is a guideline that allows you to determine if a number is divisible by another number without performing the division itself. You just need to look at the last digits or sum all the digits of the number.
Divisibility rules are helpful wherever you need to quickly understand the 'structure' of a number:
- when simplifying fractions - it's immediately clear what to divide the numerator and denominator by;
- when decomposing a number into prime factors;
- when finding the greatest common divisor and least common multiple;
- when quickly checking answers in exercises and problems.
It's important not to confuse two different questions. 'Is 3726 divisible by 9?' - this is answered by the divisibility rule: the sum of digits 3 + 7 + 2 + 6 = 18 is divisible by 9, so yes. And 'What is the result?' - this is the division itself: 3726 : 9 = 414. The rule only gives a 'yes' or 'no' answer, but it's instant and can be done mentally.
Divisibility Rules Table
It's convenient to keep all school divisibility rules in front of you in one table. Color indicates where to look: at the last digit, at the sum of digits, at the 'tail' of the last two or three digits, or at two rules at once.
| Divisible by | Divisibility Rule | Example |
|---|---|---|
| 2 | The last digit is 0, 2, 4, 6, or 8 | 3726 → 6 |
| 3 | The sum of the digits is divisible by 3 | 4512 → 4+5+1+2 = 12 |
| 4 | The last two digits form a number divisible by 4 | 1316 → 16 : 4 = 4 |
| 5 | The last digit is 0 or 5 | 1435 → 5 |
| 6 | Divisible by both 2 and 3 | 5124 → 4 and 12 |
| 8 | The last three digits form a number divisible by 8 | 3512 → 512 : 8 = 64 |
| 9 | The sum of the digits is divisible by 9 | 3726 → 3+7+2+6 = 18 |
| 10 | The last digit is 0 | 2340 → 0 |
| 11 | Alternating sum of digits from the right is divisible by 11 | 2838 → 8−3+8−2 = 11 |
| 25 | The last two digits are 00, 25, 50, or 75 | 1875 → 75 |
Divisibility Rules for 2, 5, and 10
This is the simplest group: we only look at the last digit, the others don't matter.
- numbers divisible by 2 are all even numbers - those ending in 0, 2, 4, 6, or 8;
- numbers divisible by 5 end in 0 or 5;
- numbers divisible by 10 end in 0.
The length of the number doesn't play a role: 987,654,320 is divisible by 2, 5, and 10 because it ends in zero. And 1435 is divisible by 5 (last digit is 5), but not by 2 or 10 - it's an odd number.
From this, a useful conclusion: numbers divisible by 10 are exactly those that are divisible by both 2 and 5 simultaneously.
Divisibility Rules for 3 and 9: Sum of Digits
Here, the last digit doesn't matter - the sum of all digits of the number is what counts:
- a number is divisible by 3 if the sum of its digits is divisible by 3;
- a number is divisible by 9 if the sum of its digits is divisible by 9.
If the sum turns out to be large, you can apply the rule again to the sum: for the number 899,991, the sum of digits is 8 + 9 + 9 + 9 + 9 + 1 = 45, and for 45, the sum is 4 + 5 = 9 - this means the original number is divisible by both 9 and 3.
The rules are similar but not interchangeable. Any number divisible by 9 is also divisible by 3 - because 9 itself is divisible by 3. The reverse is not always true: the sum of digits might be divisible by 3 but not by 9, and then the number only satisfies one of the two rules. Both cases are explained below.
Divisibility Rules for 4, 8, and 25: Look at the Tail of the Number
The last digit isn't enough here - you need a 'tail' of several last digits:
- a number is divisible by 4 if its last two digits form a number divisible by 4;
- a number is divisible by 8 if its last three digits form a number divisible by 8;
- a number is divisible by 25 if its last two digits are 00, 25, 50, or 75.
Why the tail is sufficient: 100 is divisible by both 4 and 25, and 1000 is divisible by 8. Therefore, everything to the left of these digits is already divisible by the required number and doesn't affect the result.
From this, an important consequence: evenness is not enough for divisibility by 4 or 8. A number can be even, but if its tail of two or three digits is not divisible, then the whole number is not divisible. The breakdown of tails is in the section below.
Divisibility Rule for 6
The number 6 doesn't have its own separate rule: 6 = 2 · 3, and the factors are coprime. Therefore, the rule is composite - a number is divisible by 6 if it is divisible by both 2 and 3 simultaneously.
Let's check 5124: the last digit is 4 - the number is even, so it's divisible by 2; the sum of digits 5 + 1 + 2 + 4 = 12 is divisible by 3. Both conditions are met → 5124 : 6 = 854.
Evenness alone is not enough: 1000 is even, but the sum of its digits is 1, which is not divisible by 3, so the number is not divisible by 6. The reverse is also not true: 4521 has a digit sum of 12, which is divisible by 3, but the number itself is odd - so it's also not divisible by 6.
Similarly, other composite rules are formed: for 12 - divisible by 3 and 4, for 15 - by 3 and 5, for 18 - by 2 and 9, for 45 - by 5 and 9.
Divisibility Rule for 11
The divisibility rule for 11 uses an alternating sum of digits. The digits are summed starting from the last one, alternating the sign: plus, minus, plus, minus. If the result is divisible by 11 (including zero), then the entire number is divisible by 11.
This same rule is often stated differently: the difference between the sum of digits in odd places and the sum of digits in even places must be divisible by 11. It's the same rule.
Examples of Divisibility Rules
Example 1. Which numbers is 7254 divisible by?
Problem. Determine which of the numbers 2, 3, 4, 5, 6, 9, 10 divide 7254.
Step 1. Divisibility by 2. The last digit is 4 - it's even. Divisible: 7254 : 2 = 3627.
Step 2. Divisibility by 5 and 10. The number ends in 4, not 0 or 5. Not divisible by 5 or 10.
Step 3. Divisibility by 3 and 9. Sum of digits: 7 + 2 + 5 + 4 = 18. It is divisible by both 3 and 9. Therefore, 7254 : 3 = 2418 and 7254 : 9 = 806.
Step 4. Divisibility by 6. The number is divisible by both 2 and 3 - therefore, it's divisible by 6: 7254 : 6 = 1209.
Step 5. Divisibility by 4. The last two digits form 54, and 54 is not divisible by 4. Not divisible.
Answer. 7254 is divisible by 2, 3, 6, and 9.
Example 2. Is 4715 divisible by 3 and 9?
Problem. Check the divisibility of the number 4715 by 3 and 9.
Step 1. Sum the digits: 4 + 7 + 1 + 5 = 17.
Step 2. Divisibility by 3. 17 is not divisible by 3 (the closest multiples are 15 and 18). Therefore, 4715 is not divisible by 3.
Step 3. Divisibility by 9. 17 is certainly not divisible by 9. Therefore, 4715 is not divisible by 9 either.
What the number IS divisible by. The last digit is 5 → the number is divisible by 5: 4715 : 5 = 943. It's not divisible by 2 or 10 - it's an odd number.
Answer. The number 4715 is divisible by neither 3 nor 9.
Example 3. Divisibility by 4 and 8: the number 6248
Problem. Check if 6248 is divisible by 4 and 8.
Step 1. Divisibility by 4. Take the last two digits: 48. The number 48 is divisible by 4 (48 : 4 = 12), so 6248 is also divisible: 6248 : 4 = 1562.
Step 2. Divisibility by 8. Take the last three digits: 248. Divide: 248 : 8 = 31 - it's divisible. Therefore, 6248 : 8 = 781.
Caution with reverse reasoning. Divisibility by 4 does not imply divisibility by 8: for the number 6244, the tail 44 is divisible by 4 (6244 : 4 = 1561), but the last three digits 244 are not divisible by 8 - so the number is not divisible by 8.
Answer. 6248 is divisible by both 4 and 8.
Example 4. Divisibility Rule for 11: the number 2926
Problem. Check if 2926 is divisible by 11.
Step 1. Write down the digits and assign alternating signs starting from the right: the last digit gets a plus, then minus, then plus.
Step 2. Calculate the alternating sum: 6 - 2 + 9 - 2 = 11.
Step 3. The result 11 is divisible by 11 → the entire number is divisible.
Check by division. 2926 : 11 = 266.
Answer. 2926 is divisible by 11.
Example 5. Simplifying a fraction using divisibility rules
Problem. Simplify the fraction 270/378.
Step 1. Calculate the sum of digits: 2 + 7 + 0 = 9 and 3 + 7 + 8 = 18. Both are divisible by 9 → both the numerator and denominator are divisible by 9.
Step 2. Simplify by 9: 270 : 9 = 30, 378 : 9 = 42. We get 30/42.
Step 3. Both numbers are even → simplify by 2: 30 : 2 = 15, 42 : 2 = 21. We get 15/21.
Step 4. The sum of digits is 6 and 3, both divisible by 3 → simplify by 3.
Answer. 270/378 = 5/7.
Common Mistakes with Divisibility Rules
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Checking divisibility by 6 based only on evenness: if a number is even, it must be divisible by 6.
Both conditions must be met: divisibility by 2 AND by 3. The number 7124 is even, but the sum of its digits is 14, which is not divisible by 3, so it's not divisible by 6. Conversely: 3471 satisfies the rule for 3 (sum is 15, 3471 : 3 = 1157), but it's odd - so it's also not divisible by 6.
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Looking at only the last digit for divisibility by 4: '8 is divisible by 4, so 9138 will work'.
For divisibility by 4, take the LAST TWO digits. For 9138, this is 38, and 38 is not divisible by 4. Correct example: for 9136, the tail is 36, 36 : 4 = 9, so 9136 : 4 = 2284. For divisibility by 8, the tail is even longer - three digits.
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Checking divisibility by 9 based on the last digit: looking for numbers ending in 9 or 0.
The last digit means nothing for divisibility by 9. Sum all the digits: for 4563, the sum is 18, which is divisible by 9 - so the number is divisible (4563 : 9 = 507), even though it ends in 3.
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Assuming that if a number is divisible by 3, it is also divisible by 9.
This works only in one direction: divisible by 9 → divisible by 3. The reverse is not true. For 8241, the sum of digits is 15: it's divisible by 3 (8241 : 3 = 2747), but not by 9.
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Using the regular sum of digits for the rule of 11, as for 3 and 9.
For 11, the sum is alternating: digits are summed from the right, changing sign every other digit. For 6479, this is 9 - 7 + 4 - 6 = 0 → it's divisible (6479 : 11 = 589). The regular sum of digits here is 26 and doesn't indicate anything.
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Confusing the divisibility rule with the actual division: answering 'yes, 18' to the question 'is 5391 divisible by 9?'.
18 is the sum of digits, an intermediate result of the rule. The quotient must be calculated separately: 5391 : 9 = 599. The rule only answers 'divisible / not divisible'.
Questions and Answers
What is a divisibility rule in simple terms?
It's a guideline that helps you determine if a number is perfectly divisible by another number without actually performing the division. Usually, you just need to look at the last digits of the number or sum all its digits.
How do I know if a number is divisible by 3?
Add up all the digits of the number. If the resulting sum is divisible by 3, then the number itself is divisible by 3. For example, for 2571, the sum of digits is 2 + 5 + 7 + 1 = 15, and 15 is divisible by 3 → 2571 : 3 = 857.
How is the divisibility rule for 3 different from the rule for 9?
The rule is the same - sum the digits, but you check the sum against different numbers: for 3, the sum must be divisible by 3; for 9, it must be divisible by 9. Therefore, any number divisible by 9 is also divisible by 3, but the reverse is not always true: for 9420, the sum of digits is 15, which is divisible by 3 (9420 : 3 = 3140), but not by 9.
What should I do if the sum of digits is large?
Apply the rule again - to the sum itself. For the number 777,771, the sum of digits is 36; for 36, the sum is 3 + 6 = 9. Since 9 is divisible by 9, the original number is also divisible by 9 (and by 3): 777,771 : 9 = 86,419.
Is there a divisibility rule for 7?
There isn't a simple rule like the sum of digits for 7, and it's not included in the standard school table of divisibility rules. There is a method: remove the last digit, double it, and subtract it from the remaining number. For example, for 343: remove 3, double it to get 6, calculate 34 - 6 = 28. Since 28 is divisible by 7, 343 is also divisible: 343 : 7 = 49.
What are the divisibility rules for 12, 15, and 45?
These are composite rules - the number must satisfy two rules simultaneously: for 12, it must be divisible by 3 and 4; for 15, by 3 and 5; for 45, by 5 and 9. It's important that the factors are coprime: the decomposition 12 = 2 · 6 is not valid, because 18 is divisible by both 2 and 6, but not by 12.
In which grade are divisibility rules studied?
In 6th grade, during math lessons: first, the rules for 10, 5, and 2, then for 3 and 9. The rules for 4, 8, 11, and 25 are usually covered additionally - they are useful for simplifying fractions and in math competitions.
