Class 8 NCERT 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: 

According to the divisibility rule of 9,the sum of all digits should be a multiple of 9

Sum of the digits of 21y5 = 2 + 1 + y + 5 = 8 + y

(8 + y) ÷ 9 = 1

8 + y = 9

y = 9 – 8 = 1

Hence, the required 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?

Solution:

According to the divisibility rule of 9, the sum of all digits should be a multiple of 9

Sum of the digits of 31z5 = 3 +1 + z + 5 = 9 + z  

9 + z = 9

z = 0

9 + z = 18

z = 9

Hence, 0 and 9 are two possible answers.

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:

Let us assume that 24x is a multiple of 3

According to the divisibility rule of 3,the sum of all digits should be a multiple of 3

Sum of the digits of 24x = 2 + 4 + x =  6 + x

6 + x = 6 if x = 0

6 + x = 9 if x = 3

6 + x = 12 if x = 6

6 + x = 15 if x = 9

Hence, x can have any of the four values

Question 4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

Solution:

According to the divisibility rule of 3,the sum of all digits should be a multiple of 3

Sum of the digits of 31z5 = 3 + 1 + z + 5 = 9 + z

9 + z = 9 if z = 0

9 + z = 12 if z =3

9 + z = 15 if z = 6

9 + z = 18 if z = 9

Hence, 0, 3, 6 and 9 are four possible values

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