NCERT Solutions for Class 8 Chapter 16 Playing with Numbers -Free PDF Download

Free PDF download of NCERT Solutions Maths Class 8 Solutions Chapter 16 – Playing with Numbers solved by Expert Maths Teachers on CoolGyan.Org. All Chapter 16 – Playing with Numbers Questions with Solutions for NCERT to help you to revise complete Syllabus and Score More marks.
Maths Revision Notes for Class 8

NCERT SOLVED

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.

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.

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.