Is there a method or algorithm to solve "in what base is the equation true" questions? I have been given some exercises in which I'm given some equation that doesn't hold in base ten, and I need to figure out in which base the equation does hold.
For example:
$\sqrt{41} = 5$ which I guessed by trial and error and concluded base 6.
However, there are some more complicated exercises and it would be cumbersome to try with every possible base. I am not sure how to approach these problems. Is there a way to solve them algorithmically?
Here are some other examples:
$$\frac{302}{20} = 12.1$$
$$9x^2+9x+4=0$$
$$\frac{x+3}{2x-7} = \frac{2x-1}{x-3}$$
 A: For the general case, you should consider what the base system means. When writing a number in base $b$, we are breaking the number down into integer powers of $b$. Using your example of base $b=6$, $5^2=4b^1+1b^0$. So we can apply thiss to your first problem to get that in base $b$,
\begin{align}
\frac{3b^2+2}{2b}&=b+2+b^{-1}\\
3b^2+2&=2b^2+4b+2\\
b^2-4b&=0\\
b(b-4)&=0
\end{align}
This equation has solutions of $b=0$ and $b=4$, but the original equation involved dividing by $b$, zero is not a valid solution. This means that the equality holds when $b=4$.
Your second equation doesn't seem to be a "find-the-base" problem. All of the constants shown are single-digits, which means that it is the same in any base where those digits make sense. It could be seen as a way to determine in what base is $994=0$, but the equation doesn't have any real solutions, so unless you are interested in complex base systems, that interpretation doesn't make sense.
The third equation also doesn't depend on base, as long as the base is $8$ or larger so that the number $7$ makes sense.
A: Remember, that the positional notation in some base $b$ makes use of (integer) powers of $b$. If you use the base as the variable, you get equations with a variable $b$ to solve for:
For example, your first exercise leads to:


*

*$\frac{3b^2+2b^0}{2b^1}=1b^1+2b^0+b^{-1}\quad|\cdot 2b$

*$3b^2+2=2b^2+4b+2\quad|-(2b^2+4b+2)$

*$b^2-4b=0$


This is a simple quadratic equation with solutions 0 and 4. Since 0 is no valid base, 4 is the answer. Hope it will help...
Update: I can't really figure out, what the idea behind the other two problems is. Since both of them only use 1 digit numbers, solutions should actually be independent of the base. Unless 4 in base 4 would be defined as 4 mod 4 = 0 or something like that. But otherwise the second example is unsolveable for a real variable $x$.
