Please have a look at the following article for an overview of Binary, Octal, Decimal & Hex values.

## Binary to decimal conversion

Consider a binary value with ** n** binary digits, where

**is the position of a digit i.e. {0 to n} from**

**'n'****or**

**LSB to MSB****.**

**Right to Left**$ d_n, ..... d_4, d_3, d_2, d_1 $

The decimal value is equal to

Decimal = $(d_n \times 2^n) + ... + (d_2 \times 2^2) + (d_1 \times 2^1) + (d_0 \times 2^0)$

### Example

## Binary to Octal conversion

Each ** octal** digit is in the range of

**. So to represent any octal number three binary bits are sufficient. For example, see the below combinations**

**[0-7]**```
000 ---> 0
001 ---> 1
010 ---> 2
011 ---> 3
100 ---> 4
101 ---> 5
110 ---> 6
111 ---> 7
```

Now, let's consider the following binary value **010100101111**

To find its equivalent octal value, make three binary bits as a group from ** right** to

**or**

**left****to**

**(LSB****as shown below.**

**MSB)****0 10**

**100**

**101**

**111**Replace this with the combinations shown above to get the octal value.

Octal =2457

## Binary to Hexadecimal

Each ** hex **digit is in the range of [0-F]. To represent a hex digit, 4 bits are required. For example, have a look at the below combinations.

```
0000 ---> 0
0001 ---> 1
0010 ---> 2
0011 ---> 3
0100 ---> 4
0101 ---> 5
0110 ---> 6
0111 ---> 7
1000 ---> 8
1001 ---> 9
1010 ---> A
1011 ---> B
1100 ---> C
1101 ---> D
1110 ---> E
1111 ---> F
```

Now, let's consider the following binary value `010100101111`

To find its equivalent hexadecimal value, make 4 binary bits as a group from** right to left** or (

**) as shown below.**

**LSB to MSB****0101 0010 1111**

Replace this with the combinations shown above to get a hexadecimal value.

= 52FHex