m^2 is divisible by 2 x 3^2, which means least value of m is 6
n^3 is divisible by 2^4 x 5, which means least value of n is 20

remainder when m*n is divided by 120 must be 0.

A is correct.
_________________

Constraint m^2/18= Integer, n^3/80=Integer
Remainder of m•n/120=?

m^2/18= m^2/(2•3^2)= (2•3)^2/(2•3^2)=integer .: m=6

n^3/80= n^3/(2^4•5)= (2^2•5)^3/(2^4•5)=Integer .:n =20

.: n•m/120 = 20•6/120—(r)—> 0

Answer A

Posted from my mobile device
_________________

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!
_________________