If m^2/18 and n^3/80 are integers, where m and n are positive integers

When \(m^2\) is divided by 18, the result is an integer. Let this integer be k. We can now form an equation as,
\(m^2\) = 18 * k.
Now, 18 = \(3^2\) * 2. Therefore, \(m^2\) = \(3^2\) * 2 * k.
Since m is a positive integer, \(m^2\) is a perfect square. For \(m^2\) to be a perfect square, smallest value of k = 2. This means \(m^2\) = \(3^2\) * \(2^2\), which also means m = 3 * 2 = 6.

When \(n^3\) is divided by 80, the result is an integer. Let this integer be x. We can now form another equation as,
\(n^3\) = 80 * x.
80 = \(2^4\) * 5. So, \(n^3\) = \(2^4\) * 5 * x.
Since n is also a positive integer, \(n^3\) is a perfect cube. For \(n^3\) to be a perfect cube, smallest value of x = \(2^2\) * \(5^2\). This means \(n^3\) = \(2^6\) * \(5^3\), which also means n = \(2^2\) * 5 = 20.

This means that the smallest value of m*n = 120. The remainder when m*n is divided by 120, is ZERO.
For bigger values of m*n also, the product will always be divisible by 120, and hence the remainder will always be zero.

The correct answer option is A.
Hope this helps!