Questions on conversion of numbers in some base to some other base is very common in competitive examination. Here in this post I present a simple technique to help you do such conversions.

First, let us understand what do we mean by number bases or systems. In our decimal number system, the rightmost position represents the “ones” column, the next position represents the “tens” column, the next position represents “hundreds”, etc. Therefore, the number 123 represents 1 hundred and 2 tens and 3 ones, whereas the number 321 represents 3 hundreds and 2 tens and 1 one.

The values of each position correspond to powers of the base of the number system. So for our decimal number system, which uses base 10, the place values correspond to powers of 10:

… 1000 100 10 1

… 10^{3 } 10^{2 } 10^{1} 10^{0}

## Binary to Decimal Conversion

Other number systems use different bases. The binary number system uses base 2, so the place values of the digits of a binary number correspond to powers of 2. For example, the value of the binary number 10011 is determined by computing the place value of each of the digits of the number:

1 0 0 1 1 the binary number

2^{4} 2^{3} 2^{2} 2^{1} 2^{0} place values

So the binary number 10011 represents the value

= (1 x 2^{4}) + (0 x 2^{3}) + (0 x 2^{2}) + (1 x 2^{1}) + (1 x 2^{0})

= 16 + 0 + 0 + 2 + 1

= 19

## Conversion from Other Number Bases to Decimals

The same principle applies to any number base. For example, the number 2132 base 5 corresponds to

2 1 3 2 number in base 5

5^{3} 5^{2} 5^{1 }5^{0} place values

So the value of the number is

(2 × 5^{3}) + (1 × 5^{2}) + (3 × 5^{1}) + (2 × 5^{0})

= (2 × 125) + (1 × 25) + (3 × 5) + (2 × 1)

= 250 + 25 + 15 + 2

= 292

This post is related to a previous post where I discussed conversion of decimal numbers into other number bases.

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