NCERT Solutions for Class 8 Maths Exercise 16.2

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NCERT Solutions for Class 8 Maths Exercise 16.2 Class 8 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 8 Maths chapter wise NCERT solution for Maths Book all the chapters can be downloaded from our website and myCBSEguide mobile app for free

NCERT solutions for Class 8 Maths Playing with Numbers Download as PDF

NCERT Solutions for Class 8 Maths Exercise 16.2

NCERT Solutions for Class 8 Maths Playing with Numbers

Class –VIII Mathematics (Ex. 16.2)
NCERT SOLUTION
1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?

Ans. Since 21y5 is a multiple of 9.

Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

 

Since 21y5 is a multiple of 9.


2. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?

Ans.  Since 31z5 is a multiple of 9.

Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

 

 

 

If

 

 

Hence 0 and 9 are two possible answers.


NCERT Solutions for Class 8 Maths Exercise 16.2

3. If 24x is a multiple of 3, where x is a digit, what is the value of x?

(Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18, … .But since x is a digit, it can only be that
6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of (four different values.)

Ans.  Since  is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

 

Since  is a digit.

     

     

    

    

Thus,  can have any of four different values.


NCERT Solutions for Class 8 Maths Exercise 16.2

4.     If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

Ans. Since 31z5 is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

Since  is a digit.

    

If  

  

If

  

If  

    

Hence 0, 3, 6 and 9 are four possible answers.

NCERT Solutions for Class 8 Maths Exercise 16.2

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