# Difficult Permutation-Combination Solved QuestionAptitude Discussion

 Q. How many factors of $2^5 × 3^6 × 5^2$ are perfect squares?
 ✖ A. $20$ ✔ B. $24$ ✖ C. $30$ ✖ D. $36$

Solution:
Option(B) is correct

Any factor of this number should be of the form $2^a × 3^b × 5^c$.

For the factor to be a perfect square $a, b, c$ have to be even.

$a$ can take values $0, 2, 4$. $b$ can take values $0,2, 4, 6$ and $c$ can take values $0,2$.

Total number of perfect squares,

$= 3 × 4 × 2$

$= \textbf{24 }$

## (3) Comment(s)

Aman
()

I can't understand thee logic behind taking the powers of a,b&c.

Thanks

AKANCHHA
()

I DONT UNDERSTAND

Rohit
()

Akansha, Its very easy.
See the number given is $2^5\times 3^6\times 5^2$, a factor is a number which divides the number fully.
So factor will have to be multiple of 2, 3 and 5 or we can say it should be of the form $2^a\times 3^b\times 5^c$.

BUT, , factor should divide the given number fully so the powers of 2, 3 and 5(i.e. a, b, c in this case) must not exceed the respective powers in number or in simpler words we have these conditions:
$a \leq 5$, $b \leq 6$ and $c \leq 2$

Now as per the questions out of all possible factors we need to select only SQUARE values for which a, b and c can take only even values.

So $a= 0,2,4$; or it can take 3 values

$b=0,2,4,6$; or it can take 4 values

$c=0,2$ or it can take 2 values

So total number of factors:

$=3\times 4\times 2$

$=\textbf{24}$