Look at several examples of rational numbers in the form $\frac{p}{q}$ $(q \neq 0)$,where $p$ and $q$ are integers with no common factors other than $1$ and having terminating decimal representations (expansions). Can you guess what property $q$ must satisfy?

(N/A) Let us examine the decimal expansion of the following terminating rational numbers:
$\frac{3}{2} = \frac{3 \times 5}{2 \times 5} = \frac{15}{10} = 1.5$ [Denominator $= 2 = 2^1$]
$\frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10} = 0.2$ [Denominator $= 5 = 5^1$]
$\frac{7}{8} = \frac{7 \times 125}{8 \times 125} = \frac{875}{1000} = 0.875$ [Denominator $= 8 = 2^3$]
$\frac{8}{125} = \frac{8 \times 8}{125 \times 8} = \frac{64}{1000} = 0.064$ [Denominator $= 125 = 5^3$]
$\frac{13}{20} = \frac{13 \times 5}{20 \times 5} = \frac{65}{100} = 0.65$ [Denominator $= 20 = 2^2 \times 5^1$]
$\frac{17}{16} = \frac{17 \times 625}{16 \times 625} = \frac{10625}{10000} = 1.0625$ [Denominator $= 16 = 2^4$]
We observe that the prime factorization of $q$ (i.e.,the denominator) has only powers of $2$,powers of $5$,or powers of both. Thus,$q$ must be of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.