# CBSE Class 8 Maths Revision Notes Chapter 16

## CBSE Class 8 Mathematics Revision Notes Chapter 16 – Playing with Numbers

In this Class 8 Mathematics chapter 16 notes, students will learn about the fun concept of playing with numbers. In addition, in this Class 8 Chapter 16 Maths notes, students will get to know the significant details of the chapter that are important for their final examination. Along with chapter 16 Mathematics Class 8 notes, Extramarks will provide students with important questions that will help them to prepare for the exams quickly. Moreover, Class 8 Mathematics notes chapter 6 will be students’  last-minute revision guide providing all the necessary information about the chapter with essential notes that make it convenient for students to remember everything clearly during the exams. These notes are based on the CBSE syllabus.

These Class 8 chapter 16 notes playing with numbers contain various types of numbers and their uses in mathematical equations. In this chapter, you will learn that there are many types of numbers in the form of whole numbers, natural numbers, integers, rational numbers, etc. This chapter talks about the test of divisibility by exploring digits.

### Access Class 8 Mathematics Chapter 16 – Playing with Numbers

Numbers in General Form

1. If a number is expressed as the sum of the products of its digits with their respective place values, it can be written in a general form.
2. ab is 10×a+b is the general form of a two-digit number.
3. 100a+10b+c is the general form of the three-digit number abc.

Games with Numbers

1. Reversing Two-digit Number:
• Decide on a two-digit number.
• To create a new number, reverse the selected digit.
• Add the initial number to this amount.
• Now divide the result by 11.
• No remainder will be left.
1. Reversing Three-digit Number:
• Decide on a three-digit number.
• To create a new number, reverse the selected digit.
• Take the smaller number out of the larger one.
• Divide the result by 99.
• No remainder will be left.
1. Forming Three-digit Numbers with Given Three-digits:
• Decide on a three-digit number.
• Using the selected number, create two additional three-digit numbers.
• Using the selected number, create two additional three-digit numbers. Then add them.
• Next, divide the result by 37.
• No remainder will be left.

Letters for Digits

Rule to remember while solving  letters for digits puzzles:

• A single digit must be used to represent each digit.
• A number’s first digit must not be zero.

Example:

The digit in column one should be Q=8 because, from Q+3, we get 1, which is a number whose one digit is 1.

Tests of Divisibility

• Divisibility by 10

When a number’s one digit is 0, it is only a multiple of 10 and, therefore, divisible by 10. For instance, 520, 950, 630, 20, etc.

• Divisibility by 5

When a number’s one digit is either 0 or 5, then only that number is a multiple of 5. That number is divisible by 5. For instance, 50, 95, 65, 20, etc.

• Divisibility by 2

When a number’s one digit is 0, 2, 4, 6, or 8, it is divisible by 2; consequently, every even number is divisible by 2. For instance: 50, 92, 64, 26, etc.

• Divisibility by 9 and 3

When the sum of a digit of a number is 3, that number is divisible by three. For instance, the number 7263 is divisible by 3 because the sum of its digits is 18, and 18 is a multiple of 2. When the sum of digits of a number is 9, that number is divisible by 9.

For instance, the number 215847 is divisible by 9 because of the sum of its digits, 27. Therefore, 215847 is also divisible by 9.

### Examples of Playing with Numbers

We must determine the values of A and B in the problem above.

Here we have A + 5 = 2 by adding the values of the one.

that means one value of the response is 2;

This means that A=7.

Hence  7 + 5 = 12. One place’s value of 12 is 2.

By applying that

3 A

+ 2 5

Becomes

B 2

3 7

+ 2 5

becomes

6 2

A has a value of 7, and B has a value of 6.

We can also find the values in multiplication problems in the same manner.

B A

* B 3

becomes

5 7 A

Here, we must discover the values of A and B.

A should have a value of 0 or 5 because the one’s digit in the expression A*3 is A.

As a result, we choose the other option, which will be true if A = 0 and B = 2.

20 * 23 = 460

Choose a different course of action if this is also incorrect.

if A = 5 and B = 2, then

25 * 23 = 575.

therefore, A = 5 and B = 2.

### Test of Divisibility in Class 8 Revision Notes Playing With Numbers

In this part of Class 8, Chapter 14, you will learn about the divisibility of 2, 3, 5, 6, 9, and 10.

Divisibility of 2

Let’s first examine the divisibility of 2:

Let’s review the natural numbers: 1, 2, 3, 4, 5, 6, 7, 8, and 9.

A whole number is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, …….

If a natural number’s last digit is 2, 4, 6, 8, or 0, it becomes an even number.

All even numbers are divisible by 2.

If the formula for the number ABC is 100*A + 10*B +C,

Here, 100 A and 10 B can be divided by two because their value in one’s place is zero. At the same time, the value of C determines if it can be divisible by 2.

At the same time, C’s value determines whether it can be divided by 2 or not.

If C is even, then ABC can be divided by two because 100*A, 10*B, and C can all be divided by two, which means that their sum can also be divided by 2.

2 can not divide ABC if C is odd or any other value, such as an integer or another number.

Example:

128 can be written as (100 * 1) + (10 * 2) + 8

= 100+20+8

Here, 100 can be divided by 2.

20 can be divided by 2.

8 can be divided by 2 as well.

Therefore, 128 can be divided by 2 as well.

Divisibility of 3

According to the mathematical principle, a number is divisible by three if the sum of its digits is a multiple of three. Example:

Let’s look at 1254 to see if it can be divided into smaller amounts.

1 + 2 + 5 + 4 = 12. 12 is a multiple of 3 here. It can therefore be divided by 3.

Divisibility of 9

The rule of mathematics that applies in this situation is that if a number’s digit sum is a multiple of 3, the number itself is divisible by 9.

Take the number 3582 into consideration. Its digit sum of 3 + 5 + 8 + 2 is 18, which can be divided by 9. 3582 is therefore also divisible by 9.

Divisibility of 5

If the last digit is either 5 or 0, then the number is divisible by 5, according to the mathematical rule in this situation.

Divisibility of 10

Here, the rule of mathematics is “if the last digit is 0, the number is divisible by 10.”

Test of Divisibility Example

Find the value of y if 21y5 is a multiple of 9 and y is a digit.

Solution:

A number’s digit sum will also be divisible by 9 if the number is divisible by 9. so 2 + 1 + y + 5 = 8 + y.

Thus, y could be 1, 10, etc.

Y = 1 because y is a single digit.

The result is 2115.

The method learned above can be used to solve puzzles.

### Why choose Extramarks?

These notes contain all the topics in the NCERT books concisely. Download the Class 8 Revision Notes for Mathematics Chapter 7 on this page to improve your exam results. Once you go through these notes, you can easily solve CBSE sample papers to further improve your mathematical ability to solve questions. These notes also contain CBSE extra questions that will help students to test their level of preparedness for the exam. .

These CBSE revision notes are made according to the CBSE syllabus. The questions in these notes are based on CBSE’s past years’ question papers. These notes contain Important questions and Formulas to help students achieve high grades in their examinations.

### 1. How can we represent numbers in a general form?

A number with 2 digits can be expressed in the form

10a + b=ab

### 2. What do you mean by a two-digit number?

A two-digit number is a number with a ones and tens place value.

### 3. What is the principle for divisibility of 5?

The principle states that if the last of a given number is either 5 or 0, then it will be divisible by 5.

### 4. What is the principle for divisibility of 10?

The principle states that if the last digit of a given number is 0, then it will be divisible by 10.