PROBLEMS ON DIVISIBILITY RULES

Problems on Divisibility Rules :

In this section, we will learn, how to solve problems using divisibility rules. 

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Problems on Divisibility Rules

Problem 1 :

Check whether 18 is divisible by 2. 

Solution :

According to the rule, if a number is an even number, then it is divisible by 2.

18 is an even number.

So, 18 is divisible by 2.

Problem 2 :

Check whether 342 is divisible by 3.

Solution :

According to the rule, if the sum of the digits in a number is a multiple of 3, then it is divisible by 3. 

Sum of the digits in 342 : 

3 + 4 + 2  =  9

9 is a multiple of 3.

So, 342 is divisible by 3.

Problem 3 :

Check whether 328 is divisible by 4.

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 divisible by 4, then it is divisible by 4.  

In the number 328, the number formed by last two digits is 28 which is divisible by 4.

So, 328 is divisible by 4. 

Problem 4 :

Check whether 105 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 105, the last digit is 5.

So, 105 is divisible 5.

Problem 5 :

Check whether 5832 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 5832 is an even number, it is divisible by 2. 

Sum of the digits in 5832 :

5 + 8 + 3 + 2  =  18

Because 18 is a multiple of 3, 5832 is divisible by 3. 

Now, it is clear that 5832 is divisible by both 2 and 3. 

So, 5832 is divisible by 6. 

Problem 6 : 

Check whether 504 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 504, twice the digit in one's place is

=  2 ⋅ 4

=  8

The number formed by the digits except the digit in one's place is

=  50

The difference between twice the digit in one's place and the number formed by the other digits is 

=  50 - 8

=  42

42 is divisible by 7.

So, the given number 504 is divisible by 7.

Problem 7 : 

Check whether 4328 is divisible by 8. 

Solution :

According to the divisibility rule for 6, 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 4328, the last three digits are not zeroes.

But, the number formed by the last three digits is 328 which is divisible by 8. 

So, the given number 4328 is divisible by 8.

Problem 8 :

Check whether 998 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 998 : 

9 + 9 + 8  =  26

26 is not a multiple of 9

So, 998 is not divisible by 9. 

Problem 9 :

Check whether 9470 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 9470, the digit in one's place is 0. 

So, it is divisible by 10.

Problem 10 :

Check whether 1782 is divisible by 11.

Solution :

According to the rule, in a number, if the sum of the digits in odd places and 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 1782, the sum of the digits in odd places :

1 + 8  =  9

In 1782, the sum of the digits in even places :

7 + 2  =  9

In 1782, the sum of the digits in odd places and sum of the digits in even places are equal.

So, 1782 is divisible by 11.

Problem 11 : 

Check whether 8520 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 8520 :

8 + 5 + 2 + 0  =  15

Because 15 is a multiple of 3, 8520 is divisible by 3. 

In 8520, the number formed by the last two digits is 20 which is divisible by 4. 

Therefore 8520 is divisible by 4. 

Now, it is clear that 8520 is divisible by both 3 and 4. 

So, 8520 is divisible by 12. 

Problem 12 :

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.  

Problem 13 :

Check whether 1458 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 1458 is an even number, it is divisible by 2. 

Sum of the digits 1458 : 

1 + 4 + 5 + 8  =  18

Because 18 is a multiple of 9, 1458 is a divisible by 9.

Now, it is clear that 1458 is divisible by both 2 and 9. 

So, 1458 is divisible by 18. 

Problem 14 :

Check whether 3500 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 3500, the last two digits are zeroes.

So, 3500 is divisible by 25.

After having gone through the stuff given above, we hope that the students would have understood, how to solve problems using divisibility rules. 

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