Class 5 - Rules of Divisibility Maths

The rules of divisibility help us find whether a given number is divisible by a particular number without actually dividing them. A number is completely divisible by another number if the remainder after division is zero. In other words, the second number divides the first number completely leaving nothing. In this article we will discuss rules of divisibility in detail along with solved examples and practice problems.

Table of Contents

• Rules of Divisibility by 2
• Rules of Divisibility by 3
• Rules of Divisibility by 4
• Rules of Divisibility by 5
• Rules of Divisibility by 6
• Rules of Divisibility by 7
• Rules of Divisibility by 8
• Rules of Divisibility by 9
• Rules of Divisibility by 10
• Solved Examples on Rules of Divisibility
  
  

Rules of Divisibility by 2

A number is divisible by 2 if the digit in the one's place of the number is 0, 2, 4, 6 or 8. The numbers, like 10, 22, 134, 9,286, 10, 928, etc. are divisible by 2.
For Example: To verify if 28 is divisble by 2 we can check if the number in ones place is even or not i.e., 8. Since 8 is even the number 28 passes the divisibility by 2.

Rules of Divisibility by 3

A number is divisible by 3 if the sum of the digits of the number is divisible by 3. In 231, 2 + 3 + 1 = 6, which is divisible by 3. Therefore, 231 is divisible by 3. For Example: To check whether 351 is divisble by 3 if the sum of the digits of the number is divisible by 3 i.e., 9. Therefore, it is the divisibile by 3.

Rules of Divisibility by 4

A number is divisible by 4 if the number formed by the last 2 digits (without changing their order) is divisible by 4. In 124, the number formed by the last two digits is 24, which is divisible by 4. Hence, 124 is divisible by 4.
Example: 844 is divisible by 4 as the number formed by last two digits is 44 which is divisible by 4

Rules of Divisibility by 5

A number is divisible by 5 if the digit in the one's place of the number is 0 or 5. The numbers, like 20, 35, 165, 200, 1,890, etc. are divisible by 5.

Rules of Divisibility by 6

A number is divisible by 6 if it is divisible by both 2 and 3. The number 84 ends with 4. Hence, it is divisible by 2. Also, 8 + 4 = 12, which is divisible by 3. Therefore, 84 is divisible by both 2 and 3. Hence, it is divisible by 6.

Rules of Divisibility by 7

Divisibility rule of 7 is to make the last digit double and subtract it from the remaining number. If the result of difference is zero or a number divisible by 7 (including negative multiples), the original number is divisible by 7.

Rules of Divisibility by 8

Divisibility rule of 8 states that a number is divisible by 8 if its last three digits are zeros or if the number formed by the last three digits is divisible by 8. 

Rules of Divisibility by 9

A number is divisible by 9 if the sum of the digits of the number is divisible by 9. In 171, 1 + 7 + 1 = 9, which is divisible by 9. Thus, 171 is divisible by 9.
Example: 333 is divisible by 9 since the sum of all the three digits is 9 and is divisible by 9.

Rules of Divisibility by 10

A number is divisible by 10 if the digit in the one's place of the number is 0. The numbers, like 40, 290, 3,510, etc. are divisible by 10.

Solved Examples on Rules of Divisibility

Example 1: Check if the following numbers are divisible by 2 or not.
i) 428
ii) 342
iii) 222

Solution: To check whether the given numbers are divisible by 2 or not we need to check if the digit at ones place is even number. Let's check each number one by one:
i) 428: The number at units place is 8 and it is an even number. So, 428 is divisible by 2.
ii) 342: The number at units place is 2 and it is an even number. So, 342 is divisible by 2.
iii) 222: The number at units place is 2 and it is an even number. So, 222 is divisible by 2.

Example 2:
Which digit should replace the ∗ in the number 36∗94 so that it becomes divisible by 9?

Solution: To determine the missing digit, use the rule for divisibility by 9: the sum of all digits must be a multiple of 9.

Sum of known digits:
3 + 6 + ∗ + 9 + 4 = 22 + ∗

Now find a value of ∗ such that (22 + ∗) is divisible by 9.
The nearest multiple of 9 greater than 22 is 27.

So,
22 + ∗ = 27
∗ = 27 − 22 = 5

Therefore, the missing digit is 5.
The number becomes 36594, which is divisible by 9.

Example 3:
Find the least and greatest possible digits that can replace ∗ in the number 25,4∗6 so that it is divisible by 4.

Solution: A number is divisible by 4 if its last two digits form a number divisible by 4.
Here, the last two digits are 4∗6, so we check values of ∗ that make the number formed by ∗6 divisible by 4.

Testing possible digits:

• If ∗ = 1, then the last two digits become 16
• 16 ÷ 4 = 4 (exactly divisible)

Thus, one valid value is ∗ = 1, making the number 25,416, which is divisible by 4.

Example 4: Verify if 1,890 is divisible by 5 or not.
Solution: Divisiblity rule of 5 states that the number in ones place should either 0 or 5. 
Since 1,890 has 0 in ones place so it is divisible by 5.