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How many factors of $2^5 × 3^6 × 5^2$ are perfect squares?









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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)


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

Please Elaborate .





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 $2^a\times 3^b\times 5^c$. Huh

BUT, Shocked, 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 $c \leq 2$ Cool

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. Huh

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:

$=3\times 4\times 2$

$=\textbf{24}$ Cool