Divisibility rules are shortcuts that help you quickly determine whether one number can be divided evenly by another. Instead of performing full division, you can use simple tests based on a number’s digits. Each rule has a logical basis, making it useful in arithmetic, fractions, and factorization. This lesson introduces key divisibility rules and how to apply them efficiently.
A number is divisible by another number if the division produces a whole number result with no remainder. For example, 18 is divisible by 6 because exactly. But 18 is not divisible by 7 because with a remainder of 4.
The number you are testing is called the dividend, and the number you are dividing by is called the divisor. When a dividend is divisible by a divisor, we say the divisor is a factor of the dividend. So 6 is a factor of 18, but 7 is not.
Divisibility rules give you a way to make this determination quickly by inspecting the digits of the number rather than carrying out the full division. This is especially valuable for large numbers where long division would be slow and error-prone.
A number is divisible by another if dividing it leaves no remainder.
divisible
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The rules for 2 and 5 are the easiest to apply because you only need to look at the ones digit, the very last digit of the number.
A number is divisible by 2 if its last digit is even: 0, 2, 4, 6, or 8. This works because every ten is divisible by 2, so only the ones digit determines whether the full number is even. For example, 346 ends in 6, which is even, so 346 is divisible by 2. The number 731 ends in 1, which is odd, so 731 is not divisible by 2.
A number is divisible by 5 if its last digit is either 0 or 5. This is because multiples of 5 always end in 0 or 5 — you can verify this by listing them: 5, 10, 15, 20, 25, and so on. So 485 is divisible by 5 because it ends in 5, while 482 is not because it ends in 2.
Check only the last digit for divisibility by 2 or 5.
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A number is divisible by 10 if its last digit is 0. This is the most intuitive rule of all. When you count in tens — 10, 20, 30, 40 — every multiple of 10 ends in zero. No multiple of 10 can end in any other digit.
You can also think of divisibility by 10 as combining the rules for 2 and 5. A number divisible by both 2 and 5 must end in a digit that satisfies both rules simultaneously. The only digit that is both even and a multiple of 5 is zero. So any number ending in 0 passes both tests at once.
For example, 3,470 is divisible by 10 because it ends in 0. The number 3,475 is not divisible by 10, even though it is divisible by 5.
A number is divisible by 10 only if its last digit is zero.
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A number is divisible by 4 if its last two digits form a number that is divisible by 4. You only need to examine the tens and ones digits, ignoring everything to the left.
The reason this works is that 100 is divisible by 4. This means any number of hundreds, thousands, or larger units is automatically divisible by 4. The only part that could cause a problem is the remaining two-digit piece at the end. So you just need to test that piece.
For example, consider 1,732. The last two digits are 32. Since exactly, 1,732 is divisible by 4. Now consider 1,738. The last two digits are 38. Since with a remainder of 2, the number 1,738 is not divisible by 4.
Check the last two digits for divisibility by 4.
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The rules for 3 and 9 are the most surprising, because they do not involve just the last digit or two. Instead, you add together all the digits of the number. If that sum is divisible by 3, the original number is divisible by 3. If the sum is divisible by 9, the original number is divisible by 9.
To test 4,281 for divisibility by 3, add its digits: . Since 15 is divisible by 3 (because ), the number 4,281 is divisible by 3. To test the same number for divisibility by 9, check whether 15 is divisible by 9. It is not (since with a remainder of 6), so 4,281 is not divisible by 9.
If the digit sum is still large, you can apply the rule again. For 99,999, the digit sum is . The digit sum of 45 is , which is divisible by 9, so 99,999 is divisible by 9. This process of re-summing is called iterating the digit sum.
Add all digits together: if the sum divides by 3 or 9, so does the original number.
A number is divisible by 6 if it is divisible by both 2 and 3. This works because 6 is the product of 2 and 3, and 2 and 3 share no common factors. So if a number contains both 2 and 3 as factors, it must contain 6 as a factor as well.
To test a number for divisibility by 6, you run two checks in sequence. First, check whether the last digit is even (the rule for 2). If the number fails this test, it is not divisible by 6 and you can stop. If it passes, check whether the digit sum is divisible by 3.
For example, test 732. The last digit is 2, which is even, so it passes the test for 2. Now compute the digit sum: . Since 12 is divisible by 3, the number passes the test for 3. Because 732 is divisible by both 2 and 3, it is divisible by 6. Now test 735. The last digit is 5, which is odd. It fails the test for 2 immediately, so 735 is not divisible by 6, even though it is divisible by 3 and 5.
Divisible by 6 means divisible by both 2 and 3.
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The true power of divisibility rules comes from applying several of them together to rapidly identify a number's factors. Given a large number, you can run through a checklist of rules and build up a picture of which numbers divide it evenly.
The general strategy for composite divisors is to factor the divisor into parts that share no common factors, then test each part separately. You have already seen this with 6 = 2 times 3. The same logic applies to other composite numbers. For instance, divisibility by 12 requires divisibility by both 4 and 3, since 12 = 4 times 3 and 4 and 3 share no common factors.
A practical application is simplifying fractions. If you want to reduce the fraction , you can use divisibility rules to find common factors quickly. Test 144 and 252 for divisibility by 2: both are even. Test both for divisibility by 3: and , both pass. So 6 divides both, and you can cancel a factor of 6 from each. Working this way is far faster than guessing factors one at a time.
Break composite divisors into coprime parts and test each separately.
Test the number for divisibility by 2, 3, 4, 5, 6, 9, and 10.
Which of the given divisors divide 5,832 evenly
Simplify the fraction by finding a common factor using divisibility rules.
A simplified form of the fraction
A number is divisible by another if division leaves no remainder.
Divisible by 2: last digit is even (0, 2, 4, 6, 8).
Divisible by 3: digit sum is divisible by 3.
Divisible by 4: last two digits form a number divisible by 4.
Divisible by 5: last digit is 0 or 5.
Divisible by 6: divisible by both 2 and 3.
Divisible by 9: digit sum is divisible by 9.
Divisible by 10: last digit is 0.
Composite divisors can be broken into coprime parts and tested separately.
Divisibility rules are a gateway into number theory. They connect directly to prime factorization, the greatest common factor, the least common multiple, and fraction simplification. Every time you use these rules, you are thinking about the deep structure of multiplication hidden inside every number.
Memorize the rules in pairs: 2 and 5 use the last digit, 3 and 9 use the digit sum, 4 uses the last two digits.
Practice on car number plates, phone numbers, or prices you see around you to make the rules automatic.
When testing a number for divisibility by 6 or 12, always test the smaller factors first to eliminate quickly.
If a digit sum is large, apply the digit sum rule again to the result until you reach a small number you recognize.
Is divisible by ? State the rule you used.
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Is divisible by ? By ?
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Is divisible by ? Show your digit sum calculation.
Test for divisibility by and by . Show your work for each test.
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A number ends in and has a digit sum of . Is it divisible by ? Explain.
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Which of the following are divisible by : , , , ?
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A three-digit number has digits , , and (in that order, so the number is ). It is divisible by . What can you conclude about and ? Give all constraints.
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Find the smallest number greater than that is divisible by , , , and simultaneously. Verify your answer using the divisibility rules.
How do you value this lesson?
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