0, 1, 2, 3, 4, 5 … etc., simply these numbers are known as whole numbers. In whole numbers there is no fractional and decimal numbers. And then there is no negative numbers. For example 6, 37 and 220 these are also whole numbers.

1, 2, 3, 4, 5 … etc., without zero the number is called counting numbers. The counting numbers also a whole number.

Whole numbers are, commonly all the integer numbers (1, 2, 3...) and it has +ive numbers and -ive numbers including 0. Yes 37 is accepted as a whole number*.*

- Addition.
- Subtraction.
- Multiplication.
- Division.
- Standard type of notation.
- Extensive form notation.
- Finding the position value of numbers.
- Rounding the numbers.
- Text the word name for the given number.

Some of the properties of whole numbers.

**a) Closure property:**

The calculations of two whole numbers are denoted as a whole number.

**Example:**

25 + 12 = 37

**b) Commutative property:**

Commutative properties has explained the whole numbers are as follows,

When X and Y are whole numbers, then it is denoted as,

X + Y = Y + X

**Example:**

30 + 7 = 7 + 30 = **37**

**c) Associative property**:

Just think a, b, c are whole numbers, then the associative property point which is,

a + (b + c) = (a + b) + c

**Example:**

3 + (15 + 19) = (3+ 15) + 19 = 37.

**d) Additive property:**

Just we take ‘x’ is a whole number, then additive property can be write as x + 0 = 0 + x = x

**Example:**

37 + 0 = 0 +37 = 37

**Problem 1:**

Add the given whole number 220 + 160.

**Solution:**

Adding the given numbers 2 2 0

1 6 0

----------------

3 8 0

----------------

This is the method to solve the arithmetic whole numbers.

**Problem 2:**

Describe the expanded form for the whole number 28643.

**Solution:-**

The given number is 28643 we have to write the given number in the expanded form.

The expanded form of the given number 28643 is

2 ten thousands + 8 thousands + 6 hundreds + 4 tens + 3 ones.

Find out the Sum of the following set of whole numbers 15, 12, 12, and -2.

**Solution:-**

We have to find out the calculation of 15, 12, 12, and -2.

By adding 15 + 12 + 12 and – 2.

15 + 12 + 12 = 39

39 – 2 = 37. Here we get 37.

** Therefore the solution is 37.**

**Problem 1:**

Divide 18 by 3

**Solution:**

6

3 )18

18

0

**Therefore the solution is 6.**

Between, if you have problem on these topics Divisibility Rules for 4, please browse expert math related websites for more help on class x cbse sample papers 2011.

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**Problem 2:**

Divide 150 by 10

**Solution:**

15

10 )150

10

50

50

0

**Therefore the solution is 15.**