  # NCERT Solutions for Class 8 Maths Chapter 16 (Ex 16.2) Playing with Numbers

## 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

 Chapter Name Playing with Numbers Chapter Chapter 16 Exercise Exercise 16.2 Class Class 8 Subject Maths NCERT Solutions Board CBSE TEXTBOOK CBSE NCERT Category NCERT Solutions

# NCERT SOLVED

1. If 21y5 is a multiple of 9, where 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 31z 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 24is a multiple of 3, where is a digit, what is the value of x?

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