# Problems on Divisibility Rules | Divisibility Rules Practice Problems PDF

Problems on Divisibility Rules will help you learn how to use divisibility tests to make your calculations much simple. Problem Solving of Divisibility Rules is quite simple if you know the tricks and apply them to the questions available. Enhance your analytical and problem-solving skills by regularly solving the divisibility rules problems here.

5th Grade Math Divisibility Rules Problems and Solutions are free to print and you can use them from any corner.Â Firstly, attempt them on your own and then verify with the solutions given here to identify the knowledge gap in the concept.

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Example 1.
Is 5128 divisible by 2,4?
Solution:
The number 5128 has 8 on its unit place which is an even number so, 5128 is divisible by 2.
In the number 5128, the last two digits are 28,Â  divisible by 4. So the original number is divisible by 4.

Example 2.
Without actual division, find if 945724 is divisible by 4 and 8?
Solution:
In the number 945724, 24 is divisible by 4. So 945724 is divisible by 4.
In the number 945724, 724 is divisible by 4. So 945724 is divisible by 4.

Example 3.
Check whether 3492 is divisible by 6?
Solution:
According to the rule, if a number is divisible by both 2 and 3, then it is divisible by 6.
Because 3492 is an even number, it is divisible by 2.
Sum of the digits in 3492 :
3+4+9+2=18
18 is a multiple of 3, So 3492 is divisible by 6.
3492 is divisible by both 2 and 3.
So 3492 is divisible by 6.

Example 4.
Check whether 338 is divisible by 7
Solution:
According to the divisibility rule for 7, in a number, if the difference between twice the digit in one’s place and the number formed by other digits is either zero or a multiple of 7, then the number is divisible by 7.
In the given number 338, twice the digit in one’s place is
= 2 â‹… 8= 16
The number formed by the digits except for the digit in one’s place is= 33
The difference between twice the digit in one’s place and the number formed by the other digits is
= 33 – 16= 17
17 is not divisible by 7.
So, the given number 338 is not divisible by 7.

Example 5.
Check whether 56248 is divisible by 8?
Solution:
According to the divisibility rule for 8, in a number, if the last three digits are zeros or the number formed by the last 3 digits is divisible by 8, then the number is divisible by 8.
In the given number 56248, the last three digits are not zeroes.
In the number 56248, the last three digits are divisible by 8.
So the number 56248 is divisible by 8.

Example 6.
Check whether 659 is divisible by 9?
Solution:
According to the rule, if the sum of the digits in a number is a multiple of 9, then it is divisible by 9.
Sum of the digits in 659 :
6 + 5 + 9 = 20
20 is not a multiple of 9
So 659 is not divisible by 9.

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Example 7.
Check whether 54280 is divisible by 10.
Solution:
According to the rule, in a number, if the digit in one’s place is 0, then it is divisible by 10.
In 54280, the digit in one’s place is 0.
So, it is divisible by 10.

Example 8.
Check whether 1452 is divisible by 11.
Solution:
According to the rule, in a number, if the sum of the digits in odd places and the sum of the digits in even places are equal or they differ by a number divisible by 11, then the number is divisible by 11.
In 1452, the sum of the digits in odd places :
1 + 5 = 6
In 1452, the sum of the digits in even places :
4 + 2 = 6
In 1452, the sum of the digits in odd places and the sum of the digits in even places are equal.
So, 1452 is divisible by 11.

Example 9.
Check whether 3612 is divisible by 12?
Solution:
According to the rule, if a number is divisible by both 3 and 4, then it is divisible by 12.
Sum of the digits in 3612
3 + 6 + 1 + 2 = 12
Because 12 is a multiple of 3, 3612 is divisible by 3.
In 3612, the number formed by the last two digits is 12 which is divisible by 4.
Therefore 3612 is divisible by 4.
Now, it is clear that 3612 is divisible by both 3 and 4.
So, 3612 is divisible by 12.

Example 10.
Check whether 41295 is divisible by 15.
Solution:
According to the rule, if a number is divisible by both 3 and 5, then it is divisible by 15.
Sum of the digits in 41295 :
4 + 1 + 2 + 9 + 5 = 21
Because 21 is a multiple of 3, 41295 is divisible by 3.
In 41295, the digit in one’s place is 5.
Therefore, 41295 is divisible by 5.
Now, it is clear that 41295 is divisible by both 3 and 5.
So, 41295 is divisible by 15.

Example 11.
Check whether 2464 is divisible by 18.
Solution:
According to the rule, if a number is divisible by both 2 and 9, then it is divisible by 18.
Because 2464 is an even number, it is divisible by 2.
Sum of the digits 2464 :
2 + 4 + 6+ 4 = 16
Because 16 is not a multiple of 9, 1458 is not divisible by 9.
So, 1458 is not divisible by 18.

Example 12.
Check whether 4500 is divisible by 25.
Solution:
According to the rule, if the last two digits of a number are zeroes or the number formed by the last two digits is a multiple of 25, then the number is divisible by 25.
In 4500, the last two digits are zeroes.
So, 4500 is divisible by 25.

Example 13.
Check whether 530 is divisible by 5?
Solution:
According to the rule, if the last digit of a number is either 0 or 5, then it is divisible by 5.
In 530, the last digit is 0.
So, 530 is divisible 5.