# NCERT Class 8 Mathematics Solutions: Chapter 16 –Playing with Numbers Exercise 16.2 Part 1

1. If is a multiple of, where is a digit, what is the value of?

Since is a multiple of .

So, according to the divisibility rule of , the sum of all the digits should be a multiple of 9.

2. If is a multiple of , where is a digit, what is the value of ?

You will find that there are two answers for the last problem. Why is this so?

Since is a multiple of .

So, according to the divisibility rule of , the sum of all the digits should be a multiple of .

If

3. If is a multiple of , where x is a digit, what is the value of ?

(Since is a multiple of , its sum of digits is a multiple of is one of these numbers: 0, 3, 6, 9, 12, 15, 18 ... But since is a digit, it can only be that .

So, or 3 or . Thus, x can have any of four different values.)

Since is a multiple of .

So, according to the divisibility rule of , the sum of all the digits should be a multiple of .

So, is digit.

So, can have any of four different values.

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

Since 31z5 is a multiple of 3.

So, according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

Since z is a digit

∴3+1+z+5=9+z

→9+z=9

→z=0

If 3+1+z+5=9+z

→9+z=12

→z=3

If 3+1+z+5=9+z

→9+z=15

→z=6

If 3+1+z+5=9+z

→9+z=18

→z=9

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