Question

please help me solve this question

Answer 1

for making it a terminating value the factors should only be 5 or 2 with their respective powers so k will be 9

Answer 2

$k = {3}^{2}$

The reason is that for getting a terminating number, we need to have the denominator in the form ${2}^{n} {5}^{m}$. Now lets substitute the value.

$\frac{k}{ {2}^{5} \times {3}^{2} } = \frac{ {3}^{2} }{ {2}^{5} \times {3}^{2} } = \frac{1}{ {2}^{5} }$

This is terminating. And we can also substitute k = ${3}^{any \: even \: number}$. Ask me if you have any doubt.