# NCERT Solutions for Class 8 Maths Exercise 16.2 ## myCBSEguide App

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NCERT solutions for Class 8 Maths Playing with Numbers ## NCERT Solutions for Class 8 Maths Playing with Numbers

###### 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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