You can Download KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2 Questions and Answers helps you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 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 multiple of 9

∴ 21y5 is divisible by 9.

So 2 + 1 + y + 5 = (8 + y) must be multiple of 9

8 + y = 9 or 18

when 8 + y = 9

⇒ y = 9 – 8

⇒ y = 1

When 8 + y = 18

⇒ y = 18 – 8

⇒ y = 10

Since y is a digit.

∴ y ≠ 10 hence 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 = (9 + z) must be multiple of 9.

9 + z = 9, 18 or 27

when 9 + z = 9

z = 9 – 9

z = 0

when 9 + z = 18

z = 18 – 9

z = 9

when 9 + z = 27

z = 27 – 9

z = 18

So values of z = 0, 9 not 18 [Since z is a digit]

Question 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).

Solution:

We have 24x is a multiple of 3

∴ 2 + 4 + x = (6 + x), is multiple of 3.

if (6 + x) = 0, 3, 6, 9, 12, 15, ……….

6 + x = 0 ⇒ x = -6

6 + x = 3 ⇒ x = 3 – 6 = -3

6 + x = 6 ⇒ x = 6 – 6 = 0

6 + x = 9 ⇒ x = 9 – 6 = 3

6 + x = 12 ⇒ x = 12 – 6 = 6

6 + x = 15 ⇒ x = 15 – 6 = 9

6 + x = 18 ⇒ x = 18 – 6 = 12

Since x is a digit

∴ x = 0, 3, 6, 9

Question 4.

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

Solution:

31z5 is a multiple of 3

If (3 + 1 + z + 5) is multiple of 3

9 + z = 0, 3, 6, 9, 12

9 + z = 0 ⇒ z = 0 – 9 = -9 (Rejected gives -ve value of z)

9 + z = 3 ⇒ z = 3 – 9 = -6 (Rejected gives -ve value of z)

9 + z = 6 ⇒ z = 6 – 9 = -3 (Rejected gives -ve value of z)

9 + z = 9 ⇒ z = 9 – 9 = 0

9 + z = 12 ⇒ z = 12 – 9 = 3

9 + z = 15 ⇒ z = 15 – 9 = 6

9 + z = 18 ⇒ z = 18 – 9 = 9

9 + z = 21 ⇒ z = 21 – 9 = 12

∴ The values of z = 0, 3, 6, 9.