If x and y are positive integers and 1620x/y^2 is the square of an odd

\(\frac{1620x}{y^2} = z^2\), with z being an odd integer(or \(2k+1\), if you fancy)

Having in mind that \(\sqrt{\frac{1620x}{y^2}}\) is an integer, we can easily see that \(√y^2 = y\) and now we need to factor \(1620x\) to find an x that makes \(\frac{1}{y}\sqrt{1620x}\) also integer:

\(1620x = 3^4*2^2*5*x\)

Now we have \(\frac{3^2*2√5x}{y}\) integer. Having in mind that √5x must also be an integer, we are looking for the smallest possible integer x (since we are also looking for the smallest xy) that makes the square root integer, which is x=5.

With that we find \(\frac{90}{y} = 2k+1 (odd)\),

So, to turn an even number, such as \(90\), into an odd one we need to divide it by another even number, which makes:

\(y=2k (even)\) , but...

Hey! We are also looking for the the smallest product for xy, so y must be the smallest even number, which is 2.

Finally,
\(x = 5\) and \(y = 2\), so \(xy = 10\)

Answer: C