Mathematics – Class 8 – Chapter 16 – Playing With Numbers – Exercise 16.2 – NCERT Exercise Solution

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

Solution:

Let, 21y5 is a multiple of 9.

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

So, 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

Since, y is a single digit number, so that the sum can be 9 only.

Hence, the value of y should be 1 only.

Hence, 8+y = 8+1 = 9.

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:

Given, 31z5 is a multiple of 9.

According to the divisibility rule of 9, we know that the sum of all the digits should be a multiple of 9.

So, 3+1+z+5 = 9+z

so, 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.

 9+0 = 9 and 9+9 = 18

Hence 0 and 9 are two possible answers.

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, 24x is a multiple of 3.

According to the divisibility rule of 3, we know that the sum of all the digits should be a multiple of 3.

So, 2+4+x = 6+x

Therefore, 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 it is given that, x is a digit, then 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.

Hence, x can have any of the four different values: 0 or 3 or 6 or 9.

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

Solution:

Given, 31z5 is a multiple of 3.

According to the divisibility rule of 3, we know that the sum of all the digits should be a multiple of 3.

so, 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.

For z = 0, 9+z = 9+0 = 9

For z = 3, 9+z = 9+3 = 12

For z = 6, 9+z = 9+6 = 15

For z = 9, 9+z = 9+9 = 18

So, 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.

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