Divisibility Rule of 6: Test, Formula and Examples

Divisibility Rule of 6 states that a number is divisible by 6 when it is divisible by both 2 and 3. Therefore, the number must be even, and the sum of its digits must be divisible by 3. This page explains the rule, shortcut method, formulas and solved divisibility by 6 examples.

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What Is the Divisibility Rule of 6?

A number is divisible by 6 if it is divisible by both 2 and 3. Thus, its last digit must be even and the sum of its digits must be divisible by 3.

Since 6 = 2 × 3 and 2 and 3 are co-prime, both divisibility conditions must be satisfied. For example, 1,452 is even, and 1 + 4 + 5 + 2 = 12, which is divisible by 3. Therefore, 1,452 is divisible by 6.

Divisibility Rule of 6 Formula & Tricks

Important Formulas

Divisibility condition
N ÷ 6 is an integer ⇔ N is divisible by 2 and 3

Check both conditions separately: the last digit must be even, and the digit sum must be divisible by 3.

Digit-sum test for 3
N = aₙaₙ₋₁...a₁a₀; N is divisible by 3 ⇔ aₙ + aₙ₋₁ + ... + a₁ + a₀ is divisible by 3

Add all digits of the number and check whether the sum is a multiple of 3.

Quick Tricks

Use the two-part shortcut

First check the last digit for divisibility by 2. Then add all digits and check divisibility by 3. The number passes the test only when both checks succeed.

Example: For 7,218, the last digit 8 is even and 7 + 2 + 1 + 8 = 18, a multiple of 3. Hence, 7,218 is divisible by 6.
Reject quickly when either condition fails

An odd last digit immediately makes the number not divisible by 6. Similarly, an even number with a digit sum that is not divisible by 3 also fails.

Example: 4,526 is even, but 4 + 5 + 2 + 6 = 17, which is not divisible by 3. Therefore, 4,526 is not divisible by 6.

Divisibility Rule of 6 Concepts

Divisibility by 2 and 3 Together

Divisibility by 6 requires simultaneous divisibility by 2 and 3.

A number is divisible by 2 when its last digit is 0, 2, 4, 6 or 8. It is divisible by 3 when the sum of its digits is a multiple of 3. Both conditions must hold for divisibility by 6.

Example: For 3,714, the last digit is 4, so it is divisible by 2. Its digit sum is 3 + 7 + 1 + 4 = 15, which is divisible by 3. Hence, 3,714 is divisible by 6.

Testing a Number Using Its Digit Sum

The digit-sum test reduces the divisibility check to a small addition.

Add every digit of the given number. If the number has an even last digit and its digit sum is 3, 6, 9, 12, 15, or any other multiple of 3, the number is divisible by 6.

Example: For 8,436, the last digit is 6 and the digit sum is 8 + 4 + 3 + 6 = 21. Since 21 is divisible by 3, 8,436 is divisible by 6.

Numbers That Fail One Condition

Failure of either the divisibility-by-2 test or the divisibility-by-3 test means the number is not divisible by 6.

An odd number cannot be divisible by 6 because every multiple of 6 is even. An even number can still fail if its digit sum is not divisible by 3.

Example: 2,358 is even, but its digit sum is 2 + 3 + 5 + 8 = 18, so it is divisible by 3 and therefore divisible by 6. In contrast, 2,356 has digit sum 16 and is not divisible by 6.

Finding a Missing Digit

For a number with a missing digit, apply the even-last-digit condition and the digit-sum condition together.

If the missing digit is in the units place, it must be even. Then choose the digit that makes the total digit sum a multiple of 3. If the missing digit is elsewhere, only the digit-sum condition is added to the existing even-last-digit condition.

Example: In 4□2, the last digit is already even. The known digit sum is 4 + 2 = 6, so □ must be 0, 3, 6 or 9 to keep the total divisible by 3.

Divisibility Rule of 6 Video Lessons

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Lesson 1 of 14 Quick Revision

Divisibility Rules for 6, 12, 15 and 18

Learn the divisibility rules for 6, 12, 15, and 18, and understand how to apply these tests to determine whether numbers are divisible by each divisor.

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Practice Divisibility Rule of 6 Questions

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1If x = √144 ÷ √36 × √49 , then what is the value of x² ?→ 2If the square of a number is increased by 56, the result is 225. find the number.→ 3The LCM of two numbers is 168, and their HCF is 14. If one of the numbers is 42, find the other number.→ 4The largest four-digit number which when divided by 6, 14 and 7 leaves remainder 5 in each case is:→ 5How many numbers among 1210, 1331, 1442, 1553 are divisible by 11?→ 6The largest four-digit number which when divided by 8, 7 and 11 leaves remainder 4 in each case is:→ 7Which of the following number is neither divisible by 3 nor by 11?→ 8Three bells ring together at 3:00 PM. They ring at intervals of 18 minutes, 24 minutes, and 32 minutes. After how much time will they ring together again?→ 9√(√400 + √25) = ?→ 10What is the largest 4-digit number that is divisible by 88?→

Divisibility Rule of 6 Quick Quiz

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Quick Revision Notes

Quick Revision: Divisibility Rule of 6

Apply both component tests because 6 is the product of 2 and 3.

  • 6 = 2 × 3.
  • A number divisible by 6 must be even.
  • Its last digit must be 0, 2, 4, 6 or 8.
  • The sum of its digits must be divisible by 3.
  • Both divisibility conditions must be satisfied.
  • An odd number cannot be divisible by 6.
  • For a missing digit, combine the last-digit test with the digit-sum test.

Divisibility Rule of 6 FAQs

What is the divisibility rule of 6?

A number is divisible by 6 if it is divisible by both 2 and 3. Its last digit must be even, and the sum of its digits must be divisible by 3.

Is 1,234 divisible by 6?

No. Although 1,234 is even, its digit sum is 1 + 2 + 3 + 4 = 10, which is not divisible by 3.

Is 7,326 divisible by 6?

Yes. Its last digit, 6, is even, and its digit sum is 7 + 3 + 2 + 6 = 18, which is divisible by 3.

Can an odd number be divisible by 6?

No. Every multiple of 6 is also a multiple of 2, so it must be even.

What is the smallest positive number divisible by 6?

The smallest positive number divisible by 6 is 6 itself. Its multiples include 12, 18, 24 and 30.

How do you find the missing digit in 52□4 so that the number is divisible by 6?

The last digit 4 already satisfies divisibility by 2. The known digit sum is 5 + 2 + 4 = 11, so the missing digit must make the total a multiple of 3. The possible digits are 1, 4 or 7.

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