# Exercise 16.2 Introduction to Graphs- NCERT Solutions Class 8

Go back to  'Playing with Numbers'

## 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

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

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

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

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 $$9.$$

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