Question

Let $3600=x\cdot y$, where $x$ and $y$ are natural numbers, then the possible pairs $(x,y)$ is

• A. $21$
• B. $23$
• C. $45$
• D. $46$

Answer 1

The correct option is C $45$

$3600=2^4\cdot 3^2\cdot 5^2$
Since, we can see that $3600$ is a perfect square, the number of ways in which $3600$ can be resolved as product of two numbers $=\frac{1}{2}\left[\right(4+1\left)\right(2+1\left)\right(2+1)+1]=23$
But here, $(x,y)$ can interchange their places.
$\therefore$ total possible pairs $(x,y)$ is $23\times 2-1=45$
$\because$given number is a perfect square, $(60,60)$ cannot be interchanged