**Class 8 Maths Chapter 16 Playing with Numbers**

**NCERT Solutions For Class 8 Maths Chapter 16 Playing With Numbers, (Maths) exam are Students are taught thru NCERT books in some of state board and CBSE Schools. As the chapter involves an end, there is an exercise provided to assist students prepare for evaluation. Students need to clear up those exercises very well because the questions withinside the very last asked from those. **

**Sometimes, students get stuck withinside the exercises and are not able to clear up all of the questions. To assist students solve all of the questions and maintain their studies with out a doubt, we have provided step by step NCERT Solutions for the students for all classes. These answers will similarly help students in scoring better marks with the assist of properly illustrated Solutions as a way to similarly assist the students and answering the questions right.**

**NCERT Solutions For Class 8 Maths Chapter 16 Playing with Numbers**

**Class 8 Maths Chapter 16 Playing with Numbers**

**Exercise 16.1**

Page No: 255

**Find the values of the letters in each of the following and give reasons for the steps involved.**

**1.**

Solution:

Say, A = 7 and we get,

7+5 = 12

In which one’s place is 2.

Therefore, A = 7

And putting 2 and carry over 1, we get

B = 6

Hence **A = 7 and B = 6**

**2.**

Solution:

If A = 5 and we get,

8+5 = 13 in which ones place is 3.

Therefore, A = 5 and carry over 1 then

B = 4 and C = 1

Hence, **A = 5, B = 4 and C = 1**

**3**.

Solution:

On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,

AxA = 6×6 = 36 in which ones place is 6.

Therefore, **A = 6**

**4.**

Solution:

Here, we observe that B = 5 so that 7+5 =12

Putting 2 at ones place and carry over 1 and A = 2, we get

2+3+1 =6

Hence **A = 2 and B =5**

**5.**

Solution:

Here on putting B = 0, we get 0x3 = 0.

And A = 5, then 5×3 =15

A = 5 and C=1

Hence **A = 5, B = 0 and C = 1**

**6.**

Solution:

On putting B = 0, we get 0x5 = 0 and A = 5, then 5×5 =25

A = 5, C = 2

Hence **A = 5, B = 0 and C =2**

**7.**

Solution:

Here product of B and 6 must be same as ones place digit as B.

6×1 = 6, 6×2 = 12, 6×3 = 18, 6×4 =24

On putting B = 4, we get the ones digit 4 and remaining two B’s value should be 44.

Therefore, for 6×7 = 42+2 =44

Hence **A = 7 and B = 4**

**8.**

Solution:

On putting B = 9, we get 9+1 = 10

Putting 0 at ones place and carry over 1, we get for A = 7

7+1+1 =9

Hence, **A = 7 and B = 9**

**9.**

Solution:

On putting B = 7, we get 7+1 =8

Now A = 4, then 4+7 =11

Putting 1 at tens place and carry over 1, we get

2+4+1 =7

Hence, **A = 4 and B = 7**

**10.**

Solution:

Putting A = 8 and B = 1, we get

8+1 =9

Now, again we add 2 + 8 =10

Tens place digit is ‘0’ and carry over 1. Now 1+6+1 = 8 =A

Hence **A = 8 and B =1**

**Exercise 16.2**

Page No: 260

**1. 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 y should be 1 only i.e. 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:

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, 9, 18, 27, and so on.

This implies, 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’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 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.

Thus, 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:

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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