Binary to decimal conversion

Consider a binary value with n binary digits, where ‘n’ is the position of a digit i.e. {0 to n} from LSB to MSB or 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 [0-7]. So to represent any octal number three binary bits are sufficient. For example, see the below combinations

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

Now, let’s consider the following binary value <strong><strong>010100101111</strong></strong>

To find its equivalent octal value, make three binary bits as a group from right to left or (LSB to MSB) as shown below.

<strong>0<strong>10</strong></strong> <strong><strong>100</strong></strong> <strong><strong>101</strong></strong> <strong><strong>111</strong></strong>

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 (LSB to MSB) as shown below.

0101 0010 1111

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

Hex = 52F

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