# NCERT Solutions For Class 8 MATH Chapter – 16 Playing with numbers Exercise – 16.2

Q1. 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   should be 1 only I.e. 8 +  y = 8 + 1 = 9

Q2. If 31z5 is 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,,18,27 and so on.

This implies , 9 + 0 = 9 and 9 + 9 = 18

Hence 0 and 9 are two possible answers.

Q3. If 24x is a multiple of 3, where x is a digit, what is the value of x ?

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 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.

Q4. 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.

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NCERT Solutions For Class 8 MATH Chapter – 16 Playing with numbers Exercise – 16.2 Q1. If 21y5 is a multiple…