# NCERT Class 8 Mathematics Solutions: Chapter 16 – Playing with Numbers Exercise 16.2 Part 1 (For CBSE, ICSE, IAS, NET, NRA 2022)

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

Answer:

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?

Answer:

Since is a multiple of .

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

If

So, are two possible answers.

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.)

Answer:

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?

Answer:

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.