TITLE: The smallest base $b$ for which the fraction $\frac{5445469}{5445468}$ has a finite number of decimal digits
QUESTION [8 upvotes]: Recently, I have found this problem:

Given the fraction $\frac{5445469}{5445468}$, find the smallest base $b\;\in\;\mathbf{N}$ such that, in base $b$, the fraction has a finite number of decimal digits.
To solve this problem, I have splitted the fraction into two terms as follows:
$$\frac{5445469}{5445468}=\frac{5445468}{5445468}+\frac{1}{5445468}=1+\frac{1}{5445468}$$
Now, the first term is a $1$, so in every base it's always $1$ as the numerator of the fraction. We have to find the minimu base for which $\frac{1}{5445468}$ has a finite number of terms.

I know the factorization of $5445468$ that is:
$$5445468=2^2\cdot3^4\cdot7^5$$
but how can we get $b$?
I think that in order to have a finite decimal representation, $5445468$ in base $b$ has to be multiple of $2$ and $5$. Is it correct?

REPLY [4 votes]: $$
\frac{1}{5445468}=
\frac{1}{2^2\cdot3^4\cdot7^5}=
\frac{2^3\cdot3^1\cdot7^0}{2^5\cdot3^5\cdot7^5}=
\frac{24}{42^5}=
(0\,.\,0:0:0:0:24)_{42}
$$
has finite representation in base $42$.
Now this fraction has finite representation in base $B$ iff
$$
\frac{1}{5445468}=\frac{A}{B^n}
$$
This implies $B^n= 5445468 A$ and so each prime divisor of $5445468$ divides $B$. Therefore, $B$ is at least $2\cdot3\cdot7=42$.