Question

How many of the following rational numbers have a terminating decimal expansion?
$\frac{41}{2^3\times 5^2},\frac{512}{2^9\times 3^2},\frac{17}{80},\frac{5}{24}$

• A. 3
• B. 2
• C. 1
• D. 4

Answer 1

The correct option is B 2

If $x=\frac{p}{q}$ is a rational number, such that p and q are co-prime and prime factorisation of q is of the $2^n{5}^m,$ where n and m are non-negative integers, then x has a decimal expansion which terminates.

Now, consider: $\frac{41}{2^3\times 5^2}$

Since, the denominator is of the form ${\text{2}}^n\times 5^n\mathrm{where}n=3\mathrm{and}m=2,$

Therefore, $\frac{41}{2^3\times 5^2}$ has a terminating decimal expansion.

Consider: $\frac{512}{2^9\times 3^2}=\frac{2^9}{2^9\times 3^2}=\frac{1}{3^2}$

Here, the denominator is not of the form $2^n\times 5^m.$

Therefore, $\frac{512}{2^9\times 3^2}$ does not have a terminating decimal expansion.

Consider: $\frac{17}{80}=\frac{17}{2^4\times 5}$

Here, the denominator is of the form of $2^m\times 5^n\mathrm{where}m=4\mathrm{and}n=1$

Therefore, $\frac{17}{80}$ has a terminating decimal expansion.

Consider: $\frac{5}{24}=\frac{5}{2^3\times 3}$

Here, the denominator is not of the form $2^m{5}^n.$
Therefore, $\frac{5}{24}$ does not have a terminating decimal expansion.

Thus, 2 out of the 4 given rational numbers have terminating decimal expansions.

Hence, the correct answer is option (b).