Haryana State Board HBSE 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.2 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 16 Playing with Numbers Exercise 16.2

Question 1.

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

Solution:

Since 21y5 is a multiple of 9.

i.e. 21y5 is divisible by 9.

Therefore, 2 + 1 + y + 5 = (8 + y) must be a multiple of 9. i.e. (8 + y) should be divisible by 9.

This is possible when 8 + y = 9 or 18

⇒ 8 + y = 9 or 8 + y = 18

y = 1 or y = 10

But, since x is a digit, therefore y = 10 is not possible.

Thus 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 ?

Solution:

Since 31z5 is divisible by 9.

∴ 3 + 1 + z + 5 = a multiples of 9.

9 + z = 9, 18 or 27

⇒ 9 + z = 9 9 + z = 18

⇒ z = 0 z = 9

9 + z = 27

z = 21

which is not a one digit number.

∴ The values of z are (0 or 9).

There are two values of z as the sum of digits is a multiple of 9. So, it may be 9, 18, 27 … any one of them.

Question 3.

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

[Hint: 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 a: 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.]

Solution:

We have 24x is a multiple of 3.

∴ 2 + 4 + x = 6 + x

is a multiple of 3.

∴ 6 + x = 0, 3, 6, 9, 12, 15, …

But, x is a digit.

∴ It can only be that

6 + x = 6 or 9, 12 or 15

∴ x = 0, 3, 6 or 9.

Question 4.

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

Solution:

Since 31z5 is a multiple of 3.

∴ The sum of the digits (3 + 1 + z + 5) is a multiple of 3.

So, 9 + z is one of these numbers :

0, 3, 6, 9, 12, …

9 + z = 9 ⇒ z = 1

[Values less than 9 gives negative result]

9 + z = 12 ⇒ z = 3

9 + 2= 15 ⇒ z = 6

9 + z = 18 ⇒ z = 9

9 + z = 21 ⇒ z = 12

which is not a one digit number.

∴ The values of z are 1, 3, 6 or 9.