Number base conversion

Question

How can I convert a number from one base, $b_1 \neq 10$ to another base $b_2 \neq 10$ without going through base $10$ i.e. $b_1\rightarrow 10 \rightarrow b_2$?

Answer 1

Short answer: You can do it, but you have to do arithmetic in base $b_1$. If you're using a computer, it's easy. If you are using pencil-and-paper, it may be easier to convert through base 10.

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The algorithm to convert a number $x$ to base $b$ is:

1. Set $n = 0$
2. Divide $x$ by $b$, yielding a quotient $q$ and a remainder $r$
3. Digit $r_n$ of the answer is $r$
4. If $q = 0$, halt; the answer is $r_{n}r_{n-1}\ldots r_0$.
5. Set $x = q$, $n = n+1$ and return to step 2

Let's say you want to convert 1e6 (base 17) to base 7.

We set $x = $ 1e6 and $n=0$. We divide $x$ by 7, yielding a quotient of 48 and a remainder of 1, so $r_0 = 1$ and return to step 2.

Now we divide 48 by 7, yielding a quotient of a and a remainder of 6, so $r_1 = 6$ and we return to step 2.

Now we divide a by 7 yielding a quotient of 1 and a remainder of 3, so $r_2 = 3$ and we return to step 2.

Now we divide 1 by 7 yielding a quotient of 0 and a remainder of 1, so $r_3 = 1$ and since $q=0$ we halt.

The answer is $1361_7$.

Answer 2

If you can do arithmetic in base $b_1$, you can use the technique of repeatedly dividing by $b_2$ and reading off the remainders in reverse order. For example to convert $261_{\text{seven}}$ to base four, you can carry out the following calculation (which is entirely in base seven):

$$\begin{align*} 261&=4\cdot50+1\\ 50&=4\cdot11+3\\ 11&=4\cdot2+0\\ 2&=4\cdot0+2 \end{align*}$$

Thus, $261_{\text{seven}}=2031_{\text{four}}$.

To see why it works, imagine that we’ve already written the number in base four. The remainder after division by four is just the unit’s (= least significant) digit, and the integer quotient is what’s left when that digit is removed.

For the reverse conversion, done entirely in base four:

$$\begin{align*} 2031&=13\cdot110+1\\ 110&=13\cdot2+12\\ 2&=13\cdot0+2 \end{align*}$$

Of course the middle digit has to be written $6$ in base seven instead of the base four $12$, but we get $2031_{\text{four}}=261_{\text{seven}}$, as expected.