Divisibility Rule of 15: Rules, Tricks and Examples

Divisibility Rule of 15 states that a number is divisible by 15 when it is divisible by both 3 and 5. This page explains the divisibility by 15 rule, the digit-sum method, the last-digit condition, shortcut checks and solved divisibility by 15 examples.

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What Is the Divisibility Rule of 15?

A number is divisible by 15 if it is divisible by both 3 and 5. Therefore, its digits must have a sum divisible by 3, and its last digit must be 0 or 5.

Since 15 = 3 × 5 and 3 and 5 are coprime, divisibility by both numbers guarantees divisibility by 15. For example, 345 has digit sum 3 + 4 + 5 = 12, which is divisible by 3, and it ends in 5. Hence, 345 is divisible by 15: 345 ÷ 15 = 23.

Divisibility Rule of 15 Formula & Tricks

Important Formulas

Divisibility condition
15 | N ⇔ (3 | N and 5 | N)

A number N is divisible by 15 exactly when it is divisible by both 3 and 5.

Test for divisibility by 3
Sum of digits of N is divisible by 3

Add all digits of the number. The sum must be divisible by 3.

Test for divisibility by 5
Last digit of N ∈ {0, 5}

The units digit must be either 0 or 5.

Quick Tricks

Check the last digit first

A number divisible by 15 must end in 0 or 5. If its last digit is 1, 2, 3, 4, 6, 7, 8 or 9, it cannot be divisible by 15.

Example: 428 ends in 8, so it is not divisible by 15 without calculating its digit sum.
Use the combined two-step test

First check whether the number ends in 0 or 5. Then add its digits and check whether the sum is divisible by 3.

Example: 1,275 ends in 5 and 1 + 2 + 7 + 5 = 15, which is divisible by 3. Therefore, 1,275 is divisible by 15.

Divisibility Rule of 15 Concepts

Two Conditions for Divisibility by 15

Both conditions must be satisfied: the number must be divisible by 3 and by 5.

Divisibility by 3 depends on the sum of the digits. Divisibility by 5 depends only on the units digit. If either condition fails, the number is not divisible by 15.

Example: 620 ends in 0, so it is divisible by 5. Its digit sum is 6 + 2 + 0 = 8, which is not divisible by 3. Hence, 620 is not divisible by 15.

Digit-Sum Method

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

A number is divisible by 3 when its digit sum is 3, 6, 9, 12, 15, and so on. This test is applied along with the last-digit test for 5.

Example: 4,815 has digit sum 4 + 8 + 1 + 5 = 18, a multiple of 3. Since it ends in 5, 4,815 is divisible by 15; 4,815 ÷ 15 = 321.

Numbers Divisible by 15

Every multiple of 15 is divisible by both 3 and 5 and ends in 0 or 5.

The multiples of 15 are 15, 30, 45, 60, 75, 90, 105, and so on. Consecutive multiples increase by 15, so their last digits alternate between 0 and 5.

Example: The numbers from 1 to 100 divisible by 15 are 15, 30, 45, 60, 75 and 90.

Finding a Missing Digit

For a number with a missing digit to be divisible by 15, its last digit must be 0 or 5 and its complete digit sum must be divisible by 3.

If the missing digit is not the units digit, first apply the units-digit condition to the given number. Then choose a digit from 0 to 9 that makes the total digit sum a multiple of 3.

Example: For 47x5, the number already ends in 5. Since 4 + 7 + 5 = 16, x must make 16 + x divisible by 3. Thus x can be 2, 5 or 8; the numbers are 4725, 4755 and 4785.

Divisibility Rule of 15 Video Lessons

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14 Lessons
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 15 Questions

Practise published questions related to this topic.

1What is the square root of 4900?→ 2Which is the largest number that divides 103, 121, 139 and 157 to leave the same remainder in each case?→ 3What is the smallest number which when increased by 29 is exactly divisible by 18, 12, 15 and 20?→ 4Find the sum of the squares of x and y, given that the number 83y547x is divisible by 72.→ 5Find the square of 19.→ 6If x = √484 ÷ √121+√64, then what is the value of x2?→ 7If the ratio of two numbers is 3 : 8, and their HCF is 6, then their LCM is:→ 8The LCM of two numbers is 168, and their HCF is 14. If one of the numbers is 42, find the other number.→ 9What is the Highest Common Factor of 15, 25 and 35?→ 10The ratio of two numbers is 7 : 9 and their Highest Common Factor is 3. What are the two numbers?→

Divisibility Rule of 15 Quick Quiz

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Quick Revision Notes

Divisibility Rule of 15: Quick Revision

Apply both the divisibility tests for 3 and 5.

  • 15 = 3 × 5, and 3 and 5 are coprime.
  • The digit sum must be divisible by 3.
  • The last digit must be 0 or 5.
  • Both conditions are required; satisfying only one is not enough.
  • To find a missing digit, make the complete digit sum a multiple of 3 while preserving the last-digit condition.
  • Examples of multiples of 15 include 15, 30, 45, 60, 75 and 90.

Divisibility Rule of 15 FAQs

What is the divisibility rule of 15?

A number is divisible by 15 if its last digit is 0 or 5 and the sum of its digits is divisible by 3.

Is every number divisible by 15 also divisible by 3 and 5?

Yes. Since 15 = 3 × 5, every multiple of 15 is divisible by both 3 and 5.

Is 735 divisible by 15?

Yes. It ends in 5, and its digit sum is 7 + 3 + 5 = 15, which is divisible by 3. Therefore, 735 ÷ 15 = 49.

Is 1,230 divisible by 15?

Yes. It ends in 0, and its digit sum is 1 + 2 + 3 + 0 = 6, which is divisible by 3. Thus, 1,230 ÷ 15 = 82.

Can a number ending in 5 fail to be divisible by 15?

Yes. It must also be divisible by 3. For example, 125 ends in 5, but its digit sum is 8, so it is not divisible by 15.

Can a number ending in 0 fail to be divisible by 15?

Yes. Its digit sum must also be divisible by 3. For example, 140 ends in 0, but 1 + 4 + 0 = 5, so it is not divisible by 15.

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