# NCERT Class 8 Maths Playing with Numbers

The chapter 16 begins with an introduction to numbers by recalling the concepts learnt in earlier classes such as natural numbers, whole numbers, integers and rational numbers and also the properties associated with them.Then the representation of numbers in general form is dealt which is followed by some games with numbers.This involves reversing the digits – two digit and three-digit number, and Letters for Digits- encoding of digits using alphabets.This is followed by Tests of Divisibility for various divisors like 10, 5, 2, 3, 6, 4, 8, 9, 11 and each on is discussed separately.

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Playing with Numbers

Exercise 16.1

## Chapter 16 Ex.16.1 Question 1

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

\[\frac{\begin{align}&3\,\,\,A\\+\;&2\,\,\,\,5\end{align}}{{\,\,\,B\,\,\,2}}\]

**Solution**

**Video Solution**

**What is known?**

Addition operation of two numbers

**What is unknown?**

Value of alphabets i.e. \(A\) and \(B.\)

**Reasoning: **

Each letter in the puzzle must stand for just one digit. Each digit must be represented by just one letter.

**Steps:**

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

\(1 + 3 + 2 = 6\)

Therefore, the addition is as follows.

\[\frac{\begin{align}&3\,\,\,\,\,7\\+ \,&2\,\,\,\,5\end{align}}{{\,\,\,\,\,6\,\,\,\,2\,}}\]

Clearly, \(B\) is \(6.\)

Hence, \(A\) and \(B\) are \(7\) and \(6\) respectively.

## Chapter 16 Ex.16.1 Question 2

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

\[\frac{\begin{align}&4\,\,\,\,\,\,A\\+ \,&9\,\,\,\,\,\,\,8\end{align}}{{\,C\,\,\,B\,\,\,\,\,3}}\]

**Solution**

**Video Solution**

**What is known?**

Addition operation of two numbers

**What is unknown?**

Value of alphabets i.e. \(A, B\) and \(C.\)

**Reasoning: **

Each letter in the puzzle must stand for just one digit. Each digit must be represented by just one letter.

**Steps:**

The addition of \(A\) and \(8\) is giving \(3\) i.e., a number whose ones digit is \(3.\) This is possible only when digit \(A\) is \(5.\) In that case, the addition of \(A\) and \(8\) will give \(13\) and thus, \(1\) will be the carry for the next step. In the next step,

\(1 + 4 + 9 = 14\)

Therefore, the addition is as follows.

\[\frac{\begin{align}&4\,\,\,\,5\\+ \,&9\,\,\,8\end{align}}{{14\,\,\,\,3}}\]

Clearly, \(B\) and \(C\) are \(4\) and \(1\) respectively.

Hence, \(A,\) \(B,\) and \(C\) are \(5,\, 4,\) and \(1\) respectively.

## Chapter 16 Ex.16.1 Question 3

Find the value of the letter in the following and give reasons for the steps involved.

\[\begin{align} \begin{array}{c}{1 \;{A}} \\ { \times\,{A}} \\ \hline 9 {A} \\ \hline\end{array} \end{align}\]

**Solution**

**Video Solution**

**Reasoning: **

Each letter in the puzzle must stand for just one digit. Each digit must be represented by just one letter.

**What is known?**

Multiplication operation of two numbers.

**What is unknown?**

Value of alphabet i.e \(A\)

**Steps:**

The multiplication of \(A\) and \(A\) gives a number whose one's digit is \(A\) again. Hence , \(A\) must be \(1\) or \(6.\)

Let \(A\) be \(1,\)

Therefore, \(A \times A = 1 \times 1 = 1 \ne 9\)

So, this is not possible for any value of \(A.\)

Hence\(,A\) must be \(6 \) only.

For \(A=6,\text{ we get }A \times A = 6 \times 6 = 36\)

and \(3\) will be a carry for the next step.

\(\therefore\) \(A\times1=6\times1+3\) (Carried on ) \(= 9 \)

\[\begin{align}\begin{array}{c}{1 \;{6}} \\ { \times\,{6}} \\ \hline 9 {6} \\ \hline\end{array}\end{align}\]

Hence, the value of \(A\) is \(6.\)

## Chapter 16 Ex.16.1 Question 4

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

\[\begin{align}{A \;{B}} \\ { +\;3\;{7}} \\ \hline 6\, {A} \\ \hline\end{align}\]

**Solution**

**Video Solution**

**What is the known?**

Addition operation of two numbers

**What is unknown?**

Value of alphabets i.e. \(A\) and \(B.\)

**Reasoning: **

**Steps:**

The addition of \(A\) and \(3\) is giving \(6.\) There can be two cases.

**(1) First step is not producing a carry **

In that case, \(A\) comes to be \(3\) as \(3 + 3 = 6.\) 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 unit’s digit of this addition comes to be \(3.\) It is possible only when \(B = 6.\) In this case,\(A = 6 + 7 = 13\) However, \(A\) is a single digit number. Hence, it is not possible.

**(2) First step is producing a carry **

In that case, \(A\) comes to be \(2\) as \(1 + 2 + 3 = 6.\) Considering the first step in which the addition of \(B\) and \(7\) is giving \(A\) (i.e., \(2\)), \(B\) should be a number such that the unit’s digit of this addition comes to be \(2.\) It is possible only when \(B=5\) and \(5 + 7 = 12.\)

\[\begin{align}{2 \;{5}} \\ { +\;3\;{7}} \\ \hline 6\, {2} \\ \hline\end{align}\]

Hence, the values of \(A\) and \(B\) are \(2\) and \(5\) respectively.

## Chapter 16 Ex.16.1 Question 5

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

\[\begin{align}{A \;{B}} \\ { \times \;\;{3}} \\ \hline C\, A\, {B} \\ \hline\end{align}\]

**Solution**

**Video Solution**

**What is known?**

Multiplication operation of two numbers

**What is unknown?**

Value of alphabets i.e. \(A, B\) and \(C.\)

**Reasoning: **

**Steps:**

The multiplication of \(3\) and \(B\) gives a number whose one’s digit is \(B\) again.

Hence, \(B\) must be \(0\) or \(5.\)

Let \(B\) be \(5.\)

Multiplication of first step \(= 3 × 5 = 15\)

\(1\) will be a carry for the next step.

We have, \(3 × A + 1 = CA\)

This is not possible for any value of \(A\).

Hence, \(B\) must be \(0\) only. If \(B = 0\), then there will be no carry for the next step.

We should obtain, \(3 \times A = CA\)

That is, the one’s digit of \(3 \times A\) should be \(A\). This is possible when \(A = 5\) or \(0.\)

However, \(A\) cannot be \(0\) as \(AB\) is a two-digit number.

Therefore, \(A\) must be \(5\) only. The multiplication is as follows.

\[\begin{align}{5 \;{0}} \\ { \times \;\;{3}} \\ \hline 1\,5\, {0} \\ \hline\end{align}\]

Hence, the values of \(A\),\(B\), and \(C\) are \(5, \,0, \) and \(1\) respectively.

## Chapter 16 Ex.16.1 Question 6

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

\[\begin{align}{A \;{B}} \\ { \times \;\;{5}} \\ \hline C\,A\, {B} \\ \hline\end{align}\]

**Solution**

**Video Solution**

**What is known?**

Multiplication operation of two numbers

**What is unknown?**

Value of alphabets i.e. \(A, \,B\) and \(C.\)

**Reasoning: **

**Steps:**

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 \(A = 2\) or \(7\)

The multiplication is as follows.

\[\frac{\begin{gathered}2\,\,\,\,5 \hfill \\\times \,\,\,\,5 \hfill \\\end{gathered} }{{1\,\,2\,\,5}}\,\,\,\,\,\,\,\,\frac{\begin{gathered}7\,\,\,\,5 \hfill \\\times \,\,\,\,5 \hfill \\\end{gathered} }{{3\,\,7\,\,5}}\]

If \(B = 0,\)

\(B \times 5 = B\)

\(0 × 5 = 0\)

There will not be any carry in this step.

In the next step, \(5 \times A = CA\)

It can happen only when \(A = 5\) or \(A = 0\)

However, \(A\) cannot be \(0\) as \(AB\) is a two-digit number.

Hence, \(A\) can be \(5\) only. The multiplication is as follows.

Hence, there are \(3\) possible values of

(i) \(5,\, 0,\) and \(2\) respectively

(ii) \(2, \,5,\) and \(1\) respectively

(iii) \(7,\, 5,\) and \(3\) respectively

## Chapter 16 Ex.16.1 Question 7

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

\[\begin{align}{A \;\;{B}} \\ { \times \;\;\,{6}} \\ \hline B\,B\, {B} \\ \hline\end{align}\]

**Solution**

**Video Solution**

**What is known?**

Multiplication operation of two numbers

**What is unknown?**

Value of alphabets i.e. \(A\) and \(B.\)

**Reasoning: **

**Steps:**

The multiplication of \(6\) and \(B\) gives a number whose one’s digit is \(B\) again.

It is possible only when

\(B = 0, \;2,\; 4,\; 6, \) or \(8\)

If \(B = 0,\) 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.

\(6{\text{A}} + 1 = {\text{BB}} = 22 \Rightarrow 6{\text{A}} = 21\) and 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.

\(6A + 3 = BB = 66 \Rightarrow 6A = 63\) and 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.

\(6{{A}} + 4 = {{BB}} = 88 \Rightarrow 6{{A}} = 84\) and hence, \(A = 14.\) However, \(A\) is a single digit number.

Therefore, this value of \(A\) is not possible.

If \(B = 4,\) then \(B \times 6 = 24\) and \(2\) will be a carry for the next step.

\(6A + 2 = BB = 44 \Rightarrow 6A = 42\) and hence, \(A = 7\)

The multiplication is as follows.

\[\begin{align}{7 \;\;{4}} \\ { \times \;\;\,{6}} \\ \hline 4\,4\, {4} \\ \hline\end{align}\]

Hence, the values of \(A\) and \(B\) are \(7\) and \(4\) respectively.

## Chapter 16 Ex.16.1 Question 8

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

\[\begin{align}{A \;\;{1}} \\ { +\;1\;\, {B}} \\ \hline B\;\; {0} \\ \hline\end{align}\]

**Solution**

**Video Solution**

**What is known?**

Addition operation of two numbers

**What is unknown?**

Value of alphabets i.e. \(A\) and \(B.\)

**Reasoning: **

**Steps:**

The addition of \(1\) and is giving \(0\) i.e., a number whose one’s digits is \(0.\) This is possible only when digit \(B\) is \(9.\) In that 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,

\[1 + A + 1 = B\]

Clearly, \(A\) is \(7\) as \(1 + 7 + 1 = 9 =B\)

Therefore, the addition is as follows.

\[\begin{align}{7 \;\;{1}} \\ { +\;1\;\, {9}} \\ \hline 9\;\; {0} \\ \hline\end{align}\]

Hence, the values of and are \(7\) and \(9\) respectively.

## Chapter 16 Ex.16.1 Question 9

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

\[\begin{align}&2\;\;A \;\;{B} \\ +&A\;\,B\;\; {1} \\ \hline &B\;\;1\;\; {8} \\ \hline\end{align}\]

**Solution**

**Video Solution**

**What is known?**

Addition operation of two numbers

**What is unknown?**

Value of alphabets i.e. \(A\) and \(B.\)

**Reasoning: **

**Steps:**

The addition of \(B\) and \(1\) is giving \(8\) i.e., a number whose one’s digits is \(8.\) This is possible only when digit \(B\) is \(7.\) In that case, the addition of \(B\) and \(1\) will give \(8.\) In the next step,

\[A + B = 1\]

Clearly, \(A\) is \(4.\)

\(4 + 7 = 11\) and \(1\) will be a carry for the next step. In the next step,

\[1 + 2 + A = B\]

\[1 + 2 + 4 = 7\]

Therefore, the addition is as follows.

\[\begin{align}&2\;\;4 \;\;{7} \\ +&4\;\,7\;\; {1} \\ \hline &7\;\;1\;\; {8} \\ \hline\end{align}\]

Hence, the values of \(A\) and \(B\) are \(4\) and \(7\) respectively.

## Chapter 16 Ex.16.1 Question 10

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

\[\begin{align}&1\;\;2 \;\;{A} \\ +\;&6\;\,A\;\; {B} \\ \hline &A\;\;0\;\; {9} \\ \hline\end{align}\]

**Solution**

**Video Solution**

**What is known?**

Addition operation of two numbers

**What is unknown?**

Value of alphabets i.e. \(A\) and \(B.\)

**Reasoning: **

**Steps:**

The addition of \(A\) and \(B\) is giving \(9\) i.e., a number whose ones digits 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, \(2 + A = 0\)

It is possible only when \(A = 8\)

\(2 + 8 = 10\) and \(1\) will be the carry for the next step.

\(1 + 1 + 6 = A\)

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.

\[\begin{align}&1\;\;2 \;\;{8} \\ +\;&6\;\,8\;\; {1} \\ \hline &8\;\;0\;\; {9} \\ \hline\end{align}\]

Hence, the values of \(A\) and \(B\) are \(8\) and \(1\) respectively.