### Playing with Numbers - Solutions 2

CBSE Class –VIII Mathematics

NCERT Solutions

CHAPTER - 16

Playing with Numbers (Ex. 16.2)

NCERT Solutions

CHAPTER - 16

Playing with Numbers (Ex. 16.2)

1. If 21

*y*5 is a multiple of 9, where*y*is a digit, what is the value of*y*?
Ans. Since 21

*y*5 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.

Since 21

*y*5 is a multiple of 9.
2. If 31

*z*5 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?
Ans. Since 31

*z*5 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.

If

Hence 0 and 9 are two possible answers.

3. If 24

*x*is a multiple of 3, where*x*is a digit, what is the value of*x*?
(Since 24

6 +

*x*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 that6 +

*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.)
Ans. Since 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.

Since is a digit.

Thus, can have any of four different values.

4. If 31

*z*5 is a multiple of 3, where z is a digit, what might be the values of z?
Ans. Since 31

*z*5 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.

Since is a digit.

If

If

If

Hence 0, 3, 6 and 9 are four possible answers.