These NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.2

Question 1.

It 21y5 is a multiple of 9, where y is a digit, what is the value of y?

Answer:

21Y5 is a multiple of 9

∴ 2 + 1 + y + 5 = 8 + y must be divisible by 9

(8 + y) should be 9, 18, 27,… etc.

Since y is a digit.

8 + y = 9

y = 9 – 8 = 1

∴ The value of y = 1.

Question 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?

Answer.

31z5 = 3 + l + z + 5 = 9 + z

31z5 is divisible by 9

∴ 9 + z must be equal to 9, 18, 27,… etc.

Z is a single digit number.

9 + z is one of these numbers.

9 + z = 9 then z = 0

9 + z=18 thenz= 18-9 = 9

∴ The value of z = 0 or 9.

Question 3.

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

Answer:

Since 24x is a multiple of 3, so sum of digits i.e. 6 + x is a multiple of 3.

6 + x is one of these numbers: 0, 3,6, 9 12, 15, 18,… But since x is a digit,

∴ 6 + x = 6 or 9 or 12 or 15

∴ x = 0, or 3 or 6 or 9

The value of x = 0 or 3 or 6 or 9

Question 4.

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

Answer:

Sum of the digits =3+l+z+5=9+z 9 + z is a multiple of 3

So, 9 + z is one of the numbers: 0, 3, 6, 9, 12, 15,…

But since z is a digit, 9 + z = 9 or 12 or 15 or 18 or 21…

∴ The Value of z = 0 or 3 or 6 or 9 or 15 or 18.