NCERT Solutions for Class 8 Maths Chapter 16 - Playing With Numbers

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

$\text{      3   A} $

$\dfrac{+\text{ }2\text{  5}}{\text{    }B\text{   2}}\text{ }$

Ans:

• The addition of $A$ and $5$ gives $2$ i.e., a number whose one’s digit is $2$. This is possible only when the digit $A$ is $7$. In this case, the addition of $A$ ($7$) and $5$ will give $12$ and thus, $1$ will be carried for the next step. 

• In the next step,

$1+3+2=6$

Therefore, the addition is as follows:

$\text{      3  7}  $ 

$\dfrac{+\text{ }2\text{  5}}{\text{    6   2}}\text{ } $ 

Clearly, $B$ is $6$.

Hence, $A$ and $B$ are $7$ and $6$ respectively. 

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

$ \text{        4    A}  $ 

$ \dfrac{+\text{ 9   8}}{\text{ C  B    3}}\text{ }  $ 

Ans: The addition of $A$ and $8$ gives $3$ i.e., a number whose one’s digit is $3$. This is possible only when digit $A$ is $5$. In this case, the addition of  $A$ and $8$ will give $13$ and thus, $1$ will be the carry for the next $1$step.

• In the next step,

$1+4+9=14$

Therefore, the addition is as follows:

$ \text{        4    5}  $ 

$ \dfrac{+\text{ 9   8}}{\text{   1 4    3}}\text{ }  $ 

Clearly, $B$and $C$ are $4$ and respectively.

Hence, $A,\text{ }B$, and $C$are $5,\text{ }4,$ and $1$ respectively.

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

$ \text{     1    A}  $ 

$ \dfrac{\text{   }\times \text{    A}}{\text{  9  A }}\text{ }  $ 

Ans:

• The multiplication of $A$ and $A$ gives a number whose one’s digit is $A$ again. Hence, $A$ must be $1$ or $6$.

• Let $A=$ is $1$.

Multiplication of first step  $1\text{ }\times \text{ }1\text{ }=\text{ }1$  ,  then there will be no carry for the next step. We should obtain, $1\text{ }\times \text{ }1\text{ }=\text{ }9$.This is not possible for value of $A=1$.

Hence, $A$ must be $6$. 

• We have, $6\text{ }\times \text{ }6\text{ }+\text{ }3=\text{ }9$,

Therefore, $A$ must be $6$. The multiplication is as follows:

$ \text{       1 A}  $ 

$ \dfrac{\times \text{      A}}{\text{    9 A }}\text{ }  $ 

• Hence, the values of $A is 6$.

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

$\text{      A   B} $

$\dfrac{+\text{ }3\text{  7}}{\text{    }6\text{   A}}\text{ }$

Ans: The addition of $A$ and $3$ is giving $6$. There can be two cases.

1. First step is not producing a carry

• In this case, $A$ comes to be $3$ as $3\text{ }+\text{ }3\text{ }=\text{ }6.$

• Considering the first step in which the addition of $B$ and $7$ is giving $A$(i.e., $3$), $B$should be a number such that the units digit of this addition comes to be $B$. It is possible only when $B\text{ }=\text{ }6$.

• In this case,$A\text{ }=\text{ }6\text{ }+\text{ }7\text{ }=\text{ }13$. However, $A$ is a single digit number. Hence, it is not possible.

2. First step is producing a carry

• In this case, $A$ comes to be $2$ as $1\text{ }+\text{ }2\text{ }+\text{ }3\text{ }=\text{ }6$.

• Considering the first step in which the addition of $B$ and $7$ is giving 

$ \text{        2    5}  $ 

$ \dfrac{+\text{    3    7}}{\text{      6     2}}\text{ }  $ 

$A$ (i.e., $2$), $B$ should be a number such that the units digit of this addition comes to be $2$. It is possible only when $B\text{ }=\text{ }5$ and $5\text{ }+\text{ }7\text{ }=\text{ }12.$

• Hence, the values of $A$ and $B$ are $2$ and $5$ respectively.

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

$ \text{     A    B}  $ 

$ \dfrac{\text{   }\times \text{    3}}{\text{   C A  B}}\text{ }  $ 

Ans:

• The multiplication of $3$ and $B$ gives a number whose one’s digit is $B$ again. Hence, $B$ must be $0$ or $5$.

• Let $B$ is $5$.

Multiplication of first step  $3\text{ }\times \text{ }5\text{ }=\text{ }15$ 

$1$ will be a carry for the next step.

• We have, $3\text{ }\times \text{ }A\text{ }+\text{ }1\text{ }=\text{ }CA$

This is not possible for any value of $A$.

Hence, $B$ must be $0$. 

If $B\text{ }=\text{ }0$, then there will be no carry for the next step. We should obtain, $3\text{ }\times \text{ }A\text{ }=\text{ }CA$

• That is, the one’s digit of $3\text{ }\times \text{ }A$should be $A$. This is possible when$A\text{ }=\text{ }5$ or $0$ . However, $A$cannot be $0$ as $AB$is a two-digit number.

• Therefore, $A$ must be $5$. The multiplication is as follows:

  $ \text{       5 0}  $ 

  $ \dfrac{\times \text{      3}}{\text{    150 }}\text{ }  $ 

• Hence, the values of $A,\text{ }B,$ and $C$ are $5,\text{ }0,$ and $1$respectively.

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

$ \text{     A    B}  $ 

$ \dfrac{\text{   }\times \text{    5}}{\text{   C A  B}}\text{ }  $ 

Ans:

• The multiplication of $B$and $5$ is giving a number whose one’s digit is $B$ again. This is possible when $B=5$ or $B=0$only.

• In case of $B=5$, the product, 

$B\times 5=5\times 5=25$

$2$ will be a carry for the next step.

• We have, $5\times A+2=CA$, which is possible for $A=2\text{ }or\text{ }7$

• The multiplication is as follows:

$ \text{     2  5}  $ 

$ \dfrac{\text{  }\times \text{   5}}{\text{   12 5}}\text{ }$  

$ \text{     7   5}  $ 

$ \dfrac{\text{  }\times \text{   5}}{\text{   3 7  5}}\text{ }$ 

• If \[B=0\],

            $ \text{     }B\times 5=B  $ 

            $ \Rightarrow \text{ }0\times 5=0$

        There will not be any carry in this step. 

• In the next step, $~5\text{ }\times \text{ }A\text{ }=\text{ }CA$.

It can happen only when $A\text{ }=\text{ }5$ or $A\text{ }=\text{ }0$

However, $A$ cannot be $0$ as $AB$ is a two-digit number. 

• Hence, $A$ can be $5$ only. The multiplication is as follows:

$ \text{     5   0}  $ 

$ \dfrac{\text{  }\times \text{   5}}{\text{   2 5  0}}\text{ }  $ 

• Hence, there are three possible values of \[A,\text{ }B,\text{ }and\text{ }C.\]

(i). $5,\text{ }0,$ and$\text{ }2$ respectively

(ii). $2,\text{ }5,$and $1$ respectively

(iii). $7,\text{ }5,$ and $3$ respectively

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

$ \text{     A   B}  $ 

$ \dfrac{\text{  }\times \text{     5}}{\text{   B B  B}}\text{ }  $ 

Ans:

The multiplication of $6$ and $B$ gives a number whose one’s digit is $B$ again. It is possible only when $B\text{ }=\text{ }0,\text{ }2,\text{ }4,\text{ }6,\text{ }or\text{ }8$

• If $B\text{ }=\text{ }0$, then the product will be $0$.

Therefore, this value of $B$ is not possible. If $B=2$, then $B\times 6=12$ and $1$ will be a carry for the next step.

 $ \text{    }6A+1=BB=22  $ 

 $ \Rightarrow 6A=21  $ 

Hence, any integer value of $A$ is not possible.

• If $B=6$, then $B\times 6=36$ and $3$ will be a carry for the next step.

$ 6A+3=BB=66  $ 

$ \Rightarrow 6A\text{ }=\text{ }63  $ 

Hence, any integer value of $A$ is not possible. 

• If $B=8$, then $B\times 6=48$and $4$ will be a carry for the next step.

$ 6A+4=BB=88  $ 

$ \Rightarrow 6A\text{ }=\text{ }84  $

Hence,$A\text{ }=\text{ }14$. However, $A$ is a single digit number. Therefore, this value of $A$ is not possible.

• If$B\text{ }=\text{ }4$, then $B\text{ }\times \text{ }6\text{ }=\text{ }24$and $2$ will be a carry for the next step.

 $ ~6A+2=BB=44  $ 

 $ \Rightarrow 6A\text{ }=\text{ }42  $ 

 Hence, $A\text{ }=\text{ }7$

• The multiplication is as follows: 

$ \text{     7   4}  $ 

$ \dfrac{\text{  }\times \text{   6}}{\text{   4 4  4}}\text{ }  $ 

Hence, the values of $A$and $B$are $7$ and $4$ respectively.

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

$ \text{     A   1}  $ 

$ \dfrac{\text{+  1   B}}{\text{   B   0}}\text{ }  $ 

Ans:

• The addition of $1$ and $B$ gives $0$ i.e., a number whose one’s digits is $0$. This is possible only when digit $B$ is $9$.

• In this case, the addition of $1$ and $B$ will give $10$ and thus, $1$ will be the carry for the next step.

• In the next step,

$1\text{ }+\text{ }A\text{ }+\text{ }1\text{ }=\text{ }B$

Clearly, $A\text{ }is\text{ }7\text{ }as\text{ }1\text{ }+\text{ }7\text{ }+\text{ }1\text{ }=\text{ }9=B$ 

• Therefore, the addition is as follows:

$ \text{     7   1}  $ 

$ \dfrac{\text{+  1   9}}{\text{   9   0}}\text{ }  $ 

Hence, the values of $A$ and $B$ are $7$ and $9$ respectively.

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

$ \text{     2 A  B}  $ 

$ \dfrac{\text{+ A  B 1}}{\text{   B  1  8}}\text{ }  $ 

Ans:

• The addition of $B$ and $1$ gives $8$ i.e., a number whose one’s digits is $8$. This is possible only when digit $B$ is $7$.

•  In this case, the addition of $B$ and $1$ will give $8$. In the next step,

 $A\text{ }+\text{ }B\text{ }=\text{ }1$

 Clearly,$A$is$\text{ }4$.

$4\text{ }+\text{ }7\text{ }=\text{ }11$ and $1$ will be a      carry for the next step. 

• In the next step, $1\text{ }+\text{ }2\text{ }+\text{ }A\text{ }=B$

$1\text{ }+\text{ }2\text{ }+\text{ }4\text{ }=7$

Therefore, the addition is as follows:

$ \text{     2  4  7}  $ 

$ \dfrac{\text{+ 4  7  1 }}{\text{   7  1  8}}\text{ }  $ 

Hence, the values of $A$ and $B$ are $4$ and $7$ respectively.

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

$ \text{     1  2  A}  $ 

$ \dfrac{\text{+ 6  A  B }}{\text{  A  0  9}}\text{ }  $ 

Ans:

• The addition of $A$ and $B$ is giving $9$ i.e., a number whose one’s digit is $9$. The sum can be $9$ only as the sum of two single digit numbers cannot be $19$. Therefore, there will not be any carry in this step.

• In the next step, $2\text{ }+\text{ }A\text{ }=\text{ }0$

It is possible only when $A\text{ }=\text{ }8$

$2\text{ }+\text{ }8\text{ }=\text{ }10$and $1$ will be the carry for the next step. 

$1\text{ }+\text{ }1\text{ }+\text{ }6\text{ }=\text{ }A$Clearly, $A$ is $8$. We know that the addition of $A$and $B$is giving $9$. As $A$ is $8$, therefore, $B$ is $1$.

• Therefore, the addition is as follows:

$ \text{     1  2  8}  $ 

$ \dfrac{\text{+ 6  8  1 }}{\text{  8  0  9}}\text{ }  $ 

Hence, the values of $A$and $B$ are $8$ and $1$ respectively.

Exercise 16.2

1. If $21y5$ is a multiple of $9$, where $y$ is a digit, what is the value of $y$? 

Ans:

• If a number is a multiple of $9$, then the sum of its digits will be divisible by $9$. 

Sum of digits of $21y5$:

$ 21y5\text{ }=\text{ }2\text{ }+\text{ }1\text{ }+\text{ }y\text{ }+\text{ }5\text{ }  $ 

        $ =\text{ }8\text{ }+\text{ }y  $

• Hence, $8\text{ }+\text{ }y$should be a multiple of $9$.

This is possible when $8\text{ }+\text{ }y$ is any one of these numbers $0,\text{ }9,\text{ }18,\text{ }27,$ and so on. 

• However, since $y$ is a single digit number, this sum can be $9$ only. Therefore, $y$ should be $1$ only.

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? 

Ans:

• If a number is a multiple of $9$, then the sum of its digits will be divisible by $9$.

• Sum of digits of $31z5$:

$31z5\text{ }=\text{ }3\text{ }+\text{ }1\text{ }+\text{ }z\text{ }+\text{ }5\text{ }=\text{ }9\text{ }+\text{ }z$

• Hence, $9\text{ }+\text{ }z$should be a multiple of $9$.

• This is possible when $9\text{ }+\text{ }z$is any one of these numbers$0,\text{ }9,\text{ }18,\text{ }27$, and so on.

• However, since $z$ is a single digit number, this sum can be either $9$ or $18$. Therefore, $z$ should be either $0$ or $9$.

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 $6\text{ }+\text{ }x$ is a multiple of $3$ ; so $6\text{ }+\text{ }x$ is one of these numbers: $0,\text{ }3,\text{ }6,\text{ }9,\text{ }12,\text{ }15,\text{ }18$…. But since $x$ is a digit, it can only be that$6\text{ }+\text{ }x\text{ }=\text{ }6\text{ }$or $9$ or $12$ or $15$. Therefore,$x\text{ }=\text{ }0\text{ }$or $3$ or $6$ or $9$. Thus, $x$ can have any of four different values)

Ans:

• Since $24x$ is a multiple of $3$, the sum of its digits is a multiple of $3$.

• Sum of digits of $24x\text{ }=\text{ }2\text{ }+\text{ }4\text{ }+\text{ }x\text{ }=\text{ }6\text{ }+\text{ }x$.  Hence, $6\text{ }+\text{ }x$is a multiple of $3$.
  

• This is possible when $6\text{ }+\text{ }x$ is any one of these numbers $0,\text{ }3,\text{ }6,\text{ }9,$and so on … Since $x$is a single digit number, the sum of the digits can be $6$or $9$ or $12$ or $15$and thus, the value of $x$ comes to $0$ or $3$ or $6$ or $9$. respectively.

• Thus, $x$ can have its value as any of the four different values $0,\text{ }3,\text{ }6,$ or $9$.

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

Ans:

• Since $31z5$ is a multiple of $3$, the sum of its digits will be a multiple of \[3\]. Hence, $3\text{ }+\text{ }1\text{ }+\text{ }z\text{ }+\text{ }5\text{ }=\text{ }9\text{ }+\text{ }z$ is a multiple of $3$.

• This is possible when $9\text{ }+\text{ }z$ is any one of the following: $0,\text{ }3,\text{ }6,\text{ }9,\text{ }12,\text{ }15,\text{ }18,$and so on …

• Since $z$ is a single digit number, the value of $9\text{ }+\text{ }z$ can only be $9,12,15$or $18$ and thus, the value of $x$ comes to $0,\text{ }3,\text{ }6,$ or $9$ respectively.

• Thus, $z$ can have its value as any one of the four different values$0,\text{ }3,\text{ }6,$ or $9$.

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers - PDF Download

By Class 6 and 7, students have learned the different types of numbers: natural numbers, integer numbers, whole numbers, and rational and irrational numbers. Students have also learned the relationship between these numbers, like finding factors and multiples between numbers. Class 8 Ch 16 Playing with numbers has been designed to teach students more about numbers in General form, fun little games that can be played with these numbers and letters for Digits. Class 8 Ch 16 also teaches students the divisibility of these numbers by different numbers. These ideas will help students in the reasoning test of divisibility. The test of divisibility includes divisibility by 10,  Divisibility by 5,  Divisibility by 2,  Divisibility by 9, and 3.

Importance of Class 8 Maths Chapter 16 Playing with Numbers NCERT Solutions

Chapter 16 of Class 8 maths is an excellent part of this syllabus where students learn different concepts related to numbers. They learn the algebraic expressions related to numbers. This chapter also introduces various puzzles to solve. Students will also find interesting number games explained in this chapter to play with. To make these concepts related to numbers easier to comprehend, focus on the NCERT solutions designed by the experts.

These NCERT solutions have been designed by following the latest CBSE Class 8 Maths syllabus to cover all the exercise questions. In fact, all the topics have been used to formulate these answers to the questions. The experts have kept the answers simple and easy to understand. It will help the students to focus on learning how to use the concepts well and precisely solve all the questions. The easier version and the organised format of the solutions will help students to proceed with their preparation for this chapter. Adding these solutions will be a huge benefit for the students to develop their mathematical concepts.

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Now Here is What this Chapter Includes:

There are a total of 9 sub-topics included in this chapter. The first one is an introduction in which fundamental concepts about the number and their algebra are explained.

 In the second topic, numbers in general form consist of different types of numbers and their general form. The third topic is Games with Numbers. In this, there are exciting puzzles and games which students will enjoy while solving.

The Fourth topic is Letters for digits. This fourth topic is an interesting one; in this, the student learns how to find the missing number. In this addition or subtraction will be given with one or two amounts replaced with words. And students have to find correct numbers for that word.

After that comes subtopics on the divisibility test, candidates here learn about How to identify if a certain number is divisible by 10 or 5 or 2 or 3 and 9. These Subtopics are Divisibility by 10, Divisibility by 5, Divisibility by 2, Divisibility by 9, and 3.

All questions and answers on each subtopic are covered in this Vedantu solution for Class 8th Maths Chapter 16 Playing with Number.

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Vedantu provides the detailed and step by step answers to all the questions of NCERT solutions for Class 8 maths chapter playing with numbers as per NCERT CBSE Book Guidelines, which will help you to explain the concepts of numbers and their divisibility asked in the question paper. Maths can be a very intimidating and challenging subject for students and, if not understood fully, will create fear in students regarding the subject lifelong. Finding and knowing the solution to a problem can be only one part of studying maths. The other and essential part is knowing how to arrive at a particular solution and understanding the significance of the steps used. It is an essential factor which can help you to score more marks in the exams. Our expert Maths teachers give one hundred percent accurate solutions to the questions given in Class 8 maths playing with numbers Chapter 16 textbooks per CBSE Board guidelines from the latest NCERT book for Class 8 maths. We also provide sample problems that will help you to tackle new questions or situations easily during exams. The significant benefits of opting playing with numbers Class 8 NCERT solutions are mentioned below:

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Based on the instructions and directions of NCERT and CBSE Board, you can learn about your chapter in detail using the solutions provided by the Vedantu. Creatively write all your answers utilising the solutions of 8th standard Maths Chapter 16 playing with numbers. Once you get started your learning with Vedantu, you can easily download the free PDF solution of Class 8 Maths Chapter 16 playing with numbers. We offer a user-friendly interface for you to search for the relevant materials easily using the keywords. All the answers are written exclusively by the professionals for children to grasp quickly and their understanding of the numbers. It also helps them master the concepts of different numbers and their divisibility and lay the foundation for understanding future complex topics. New exercises, fun games, activities, concepts, solved unsolved problems incorporated in Vedantu generates intrigue in students and helps students delve into the sea of Playing with numbers and Maths. Vedantu solutions not only help a student in scoring higher marks but lays the groundwork for understanding the concepts so that a student can use the knowledge gained here in real-life problems and future complex syllabus. Other than given examples, you should also practise all the solved examples from the textbook to clear the concepts.

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We Cover all Exercises in the chapter given below:-

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Get the free PDF version of these solutions and complete your study material for this chapter. Check the answering formats and solutions designed by the experts for these questions and fortify your preparation level. Learn how to use the concepts of this chapter well to precisely answer such questions during an exam and stay ahead of the competition.