# Exercise 16.2 Introduction to Graphs- NCERT Solutions Class 8

Playing with Numbers

Exercise 16.2

## Chapter 16 Ex.16.2 Question 1

If \(21y5\) is a multiple of \(9,\) where \(y\) is a digit, what is the value of \(y\)?

**Solution**

**Video Solution**

**What is known?**

A puzzled number

**What is unknown?**

Value of the alphabet i.e. \(y.\)

**Reasoning: **

If the sum of all digits of a number is equal to \(9,\) then the number is a multiple of \(9.\)

**Steps:**

If a number is a multiple of \(9,\) then the sum of its digits will be divisible by \(9.\)

Sum of digits of

\[21y5 = 2 + 1 + y + 5 = 8 + y\]

Hence, \(8 + y\) should be a multiple 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.

\[\begin{align} 8 + y &= 9\\ y &= 9 - 8\\ y &= 1 \end{align}\]

Therefore, \(y\) should be \(1\) only.

## Chapter 16 Ex.16.2 Question 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**

**Video Solution**

**What is known?**

A puzzled number

**What is unknown?**

Value of the alphabet i.e. \(y.\)

**Reasoning: **

If the sum of all digits of a number is equal to \(9,\) then the number is a multiple of \(9.\)

**Steps:**

If a number is a multiple of \(9,\) then the sum of its digits will be divisible by \(9.\)

Sum of digits of

\[31z\,5 = 3 + 1 + z + 5 = 9 + z\]

Hence, \(9 + z\) should be a multiple of \(9.\)

This is possible, when \(9 + z\) is any one of these numbers \(0,\; 9,\; 18,\; 27,\) and so on ...

However, since \(z\) is a single digit number, this sum can be either \(9\) or \(18.\)

\(9+z =9\\z=9-9\\z=0\)

Therefore, \(z\) should be either \(0\) or \(9.\)

## Chapter 16 Ex.16.2 Question 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**

**Video Solution**

**What is known?**

A puzzled number

**What is unknown?**

Value of the alphabet i.e. \(x.\)

**Reasoning: **

If the sum of the digits of a number is divisible by \(3,\) then the given number is a multiple of \(3.\)

**Steps:**

Since \(24x\) is a multiple of \(3,\) the sum of its digits is a multiple of \(3.\)

Sum of digits of

\(24x = 2 + 4 + x = 6+ x\)

Hence, \(6+x\) is a multiple of \(3.\)

This is possible when \(6+x\) is any one of these numbers \(0, \;3,\; 6,\; 9,\) and so on ...

\[\begin{align}\text{For},&\\&6 + x = 0 \Rightarrow x = 0 - 6 = - 6\\\text{For},&\\&6 + x = 3 \Rightarrow x = 3 - 6 = - 3\\\text{For},&\\&6 + x = 6 \Rightarrow x = 6 - 6 = 0\\\text{For},&\\&6 + x= 9 \Rightarrow x = 9 - 6 = 3\\\text{For},&\\&6+x =12 \Rightarrow x=12-6=6\\&\dots \text{so on} \end{align}\]

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 comes to \(0\) or \(3\) or \(6\) or \(9\) respectively.

Thus, \(x\) can have its value as any of the four different values \(0,\; 3,\; 6,\;\) or \(9.\)

## Chapter 16 Ex.16.2 Question 4

If \(31z5\) is a multiple of \(3,\) where \(z\) is a digit, what might be the values of \(z\)?

**Solution**

**Video Solution**

**What is known?**

A puzzled number

**What is unknown?**

Value of the alphabet i.e. \(z.\)

**Reasoning: **

If sum of the digits of a number is divisible by \(3,\) then the given number is a multiple of \(3.\)

**Steps:**

Since \(31z5\) is a multiple of \(3,\) the sum of its digits will be a multiple of \(3.\)

That is, \(3 + 1 + z + 5 = 9 + z\) is a multiple of \(3.\)

This is possible when \(9 + z\) is any one of \(0,\;3,\; 6, \;9,\; 12,\; 15,\; 18,\) and so on ...

\[\begin{align} \text{For},&\\&9 + z = 0 \Rightarrow z = 0 - 9 = - 9\\ \text{For},&\\&9 + z = 3 \Rightarrow z = 3 - 9 = - 6\\ \text{For},&\\&9 + z = 6 \Rightarrow z = 6 - 9 = - 3\\ \text{For},&\\&9 + z = 9 \Rightarrow z = 9 - 9 = 0\\ \text{For},&\\&9 + z = 12 \Rightarrow z = 12 - 9 = 3\\ \text{For},&\\&9 + z = 15 \Rightarrow z = 15 - 9 = 6\\ \text{For},&\\&9 + z = 18 \Rightarrow z = 18 - 9 = 9 \end{align}\]

Since \(z\) is a single digit number, the value of \(9 + z\) can only be \(9\) or \(12\) or \(15\) or \(18\) and thus, the value of \(z\) comes to \(0\) or \(3\) or \(6\) or \(9\) respectively.

Thus, \(z\) can have its value as any one of the four different values \(0, \;3,\; 6,\;\) or \(9.\)