Number base conversion 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. $b_1\rightarrow 10 \rightarrow b_2$?
 A: 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.

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


*

*Set $n = 0$

*Divide $x$ by $b$, yielding a quotient $q$ and a remainder $r$

*Digit $r_n$ of the answer is $r$

*If $q = 0$, halt; the answer is $r_{n}r_{n-1}\ldots r_0$.

*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 $1361_7$.
A: 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.
