So I've looked at the examples $\frac{1}{192}$ where we needed to find a base which gave a terminating expansion with precisely 2 digits after the point, and $\frac{22}{675}$ where the digits after the point weren't specified. (Both fractions are in base 10 to start with.)
For $\frac{1}{192}$ the base was 24 and for $\frac{22}{675}$ it was 15 but I don't understand how they got to those answers.
Could somebody please explain or give me a generalised method to do it?
Thank you!
On
Look at $192=3\times 64 = 3^1\times 2^6$
$\cfrac 1{192}$ will have a terminating decimal in any base which is divisible by $6=2\times 3$ because $192 \times 3^5=6^6$ so that $\cfrac 1{192}=\cfrac{3^5}{6^6}$ and further note that this $=\cfrac {3}{2^63^2}=\cfrac {3}{24^2}$
Here I have cancelled what I can whilst leaving a square in the denominator (everything to an even power) - this is because I wanted a result with two digits. If I were looking for three digits I'd go for $$\frac {3^2}{2^63^3}=\frac 9{12^3}$$ so I would have a cube (the cube of the base) in the denominator.
The problem of finding the smallest $b$ such that $\frac1{192}$ can be written in base $b$ with exactly $2$ digits is the problem of findint the smallest $b$ such that $\frac1{192}$ and be written as $\frac\alpha b+\frac\beta{b^2}$ with $\alpha,\beta\in\{0,1,\ldots,b-1\}$ and $\beta>0$. But $\frac\alpha b+\frac\beta{b^2}=\frac{\alpha b+\beta}{b^2}$. Now, how can we write$$\frac{\alpha b+\beta}{b^2}=\frac1{192}?$$This implies that $192\mid b^2$. Since $192=2^6\times3$, the smallest perfect square which is a multiple of $192$ is $2^6\times3^2=576=24^2$. So, since$$\frac3{24^2}=\frac3{576}=\frac1{192},$$you take $b=24$.
Can you solve the other problem now?