NCERT Solutions Class 8th Maths Chapter – 16 Playing with Numbers Exercise 16.2

Ncert Solution Class 8th (Chapter – 16) Exercise – 16.2 Question . 1

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

Solution: Suppose 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.
That is, 2+1+y+5 = 8+y
Therefore, 8+y is a factor of 9.
This is possible when 8+y is any one of these numbers 0, 9, 18, 27, and so on
However, since y is a single digit number, this sum can be 9 only.
Therefore, the value of y should be 1 only i.e. 8+y = 8+1 = 9.

Ncert Solution Class 8th (Chapter – 16) Exercise – 16.2 Question . 2

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 a multiple of 9.
Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.
3+1+z+5 = 9+z
Therefore, 9+z is a multiple of 9
This is only possible when 9+z is any one of these numbers: 0, 9, 18, 27, and so on.
This implies, 9+0 = 9 and 9+9 = 18
Hence 0 and 9 are two possible answers.

Ncert Solution Class 8th (Chapter – 16) Exercise – 16.2 Question . 3

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’s say, 24x is a multiple of 3.
Then, according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
2+4+x = 6+x
So, 6+x is a multiple of 3, and 6+x is one of these numbers: 0, 3, 6, 9, 12, 15, 18 and so on.
Since, x is a digit, the value of x will be either 0 or 3 or 6 or 9, and the sum of the digits can be 6 or 9 or 12 or 15 respectively.
Thus, x can have any of the four different values: 0 or 3 or 6 or 9.

Ncert Solution Class 8th (Chapter – 16) Exercise – 16.2 Question . 4

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.
Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
That is, 3+1+z+5 = 9+z
Therefore, 9+z is a multiple of 3.
This is possible when the value of 9+z is any of the values: 0, 3, 6, 9, 12, 15, and so on.
At z = 0, 9+z = 9+0 = 9
At z = 3, 9+z = 9+3 = 12
At z = 6, 9+z = 9+6 = 15
At z = 9, 9+z = 9+9 = 18
The value of 9+z can be 9 or 12 or 15 or 18.
Hence 0, 3, 6 or 9 are four possible answers for z.