Answer
Let $r=\dfrac{a}{b}$ in the lowest terms, which gives finite decimal expansion with $N$ non-zero decimal places. Also, $a,b$ are integers with no common divisors.
Now, $10^{N}\cdot r$ will be an integer, or, $10^{N}\cdot \dfrac{a}{b}=\dfrac{10^{N}\cdot a}{b}$ is an integer, which means $b$ completely divide $10^{N}\cdot a$. But $b$ does not divide $a$ as both are co-prime, so, $b$ divides $10^{N}=2^{N}\cdot 5^{N}$. Thus, $b$ must have prime factors $2$ and $5$, i.e., $b=2^{n}\cdot 5^{m}$, where $0\leq n,m \leq N$ and both are integers. Hence, proved.
Work Step by Step
The results to be used are as follows:
(i) If a real number, $r$ has finite decimal expansion with $N$ non-zero decimal places, then, $10^{N}\cdot r$ is an integer.
(ii) If some number, $c$ divides $a\cdot b$ but $c$ and $b$ are co-prime,i.e., have no common divisors, then, $c$ divides $a$.
(iii) If a number $a$ divides $n=p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}}\cdots p_{k}^{a_{k}}$, where $p_{i}$'s are prime factors for $n$, then, $a$ will also have same prime divisors but multiplicity can be less than or equal to multiplicity in prime factorization for $n$. For instance, if $n=2^{6}\cdot 5^{6}$, and $a$ divides $n$, then, $a=2^{n}\cdot 5^{m}$, where $0\leq n\leq 6$ and $0\leq m \leq 6$.
