# Base N to Base N Conversion

mepcotterell
I know how to convert from any base to any other base but to do so I usually have to go through base 10 with the exception of base 2 to base 16 and vise-versa. I was wondering if you guys know an easy way to convert bases?

The way I usually do it is using the base notation to get base 10 then the remainder method to finish the conversion...

For example to go from base 2 to 5 I would do this...

1011b2 to b10

1*2^0 + 1*2^1 + 0*2^2 + 1*2^3 = 1 + 2 + 0 + 8 = 11

11b10 to b5

11 / 5 = 2 R 1
2 / 5 = 0 R 2

= 21b5

Last edited:

Homework Helper
If you're converting between two bases that are both powers of the same number, then it's possible to do it in 'chunks' because there are locations that have corresponding 'orders'. Consider, for example that the digits base 4 are:
1, 4, 16, 64, and so on
while the digits base 2 are
1,2,4,8,16,32,64 and so on
So it's possible to do 'local' conversion:
Effectively a number base for can be thought of as a polynomial
$$\sum a_i 4^i$$
but that's
$$\sum a_i \left(2^2\right)^i$$
so
$\sum a_i 2^{2i}$
and, since $0\leq a_i \leq 3$ we have
$$a_i=b_{2i}*2^0+b_{2i+1}*2^i$$
so the sum is
$\sum (b_{2i}*2^0+b_{2i+1}*2^i)2^{2i}[/tex] distribute through [itex]\sum b_{2i}^{2i}+b_{2i+1}2^{2i+1}$
$\sum b_{i}2^i$
It's a little bit more complicated if you look at, for example, base $9=3^2$ and base $27=3^3$:
$$\sum b_{3i} 9^i + b_{3i+1} 9^{3i+1} + b_{3i+2}9^{3i+2}=\sum c_{2j} 27^{2j} + c_{2j+1} 27^{2j+1}[/itex] (This is similar to converting from base 9 to base 9^3=729=27^2 to base 27) Generally, there is no need to convert numbers to base 10 in order to do base conversion, but, since we're familiar with arithmetic in base 10, that's our normal tendency. Similarly, computers, when doing arithmetic computers typically convert into and out of a base that's some power of two because computers can operate on binary (or hexidecimal) numbers more efficiently. This will look a bit strange, but consider your example: Convert 1011 (base 2) to base 5 Now 2^1=1 (base 5) 2^2=4 (base 5) 2^3=13 (base 5) 2^4=31 (base 5) Then: [tex]1101 (base 2) = 1*2^4+1*2^3+0*2^2+1*2^0 (base 5) = 1*31 + 1* 13 + 0 * 4 + 1 * 1 (base 5)$$
$$=31+13+1 (base 5)=44+1 (base 5)=100 (base 5)$$