If \(x\) and \(y\) are positive integers and

1620 = (4)(405) = (4)(81)(5) = (2^2)(3^4)(5)

So if (2^2)(3^4)(5)x / y^2 is going to be odd, the 2^2 needs to cancel, and y^2 needs to be divisible by 2^2, so y needs to be divisible by 2. If we're going to end up with a perfect square, we'll need an even exponent on the '5' after we multiply by x, so x needs to be divisible by 5. Since we want the minimum values of x and y, those values are 2 and 5 (we'll then get the odd square 3^4*5^2 = 45^2), and their smallest possible product is 10.

BTGmoderatorLU wrote:Source: Manhattan Prep

If \(x\) and \(y\) are positive integers and \(\frac{1620x}{y^2}\) is the square of an odd integer, what is the smallest possible value of \(xy\)?

A. 1
B. 8
C. 10
D. 15
E. 28

The OA is C

In order for 1620x/y^2 to be a perfect square, 1620x has to be a perfect square also (since y^2 is a perfect square).

Since 1620 = 162 x 10 = 81 x 2 x 10 = 2^2 x 3^4 x 5^1, we see that in order for 1620x to be the smallest perfect square, x must be 5. Now, if y^2 = 2^2, then 1620x/y^2 = 3^4 * 5^2 is indeed the square of an odd integer. So we see that y = 2 and therefore, xy = 5 * 2 = 10.

Answer: C