Suppose \(N=2^2 \times 3^2 \times 5^2 \times k\), where \(k\) is a positive integer. If \(N\) is divisible by \(2025\), then what is the smallest possible value for \(k\)?
First we note that \(2025=3^4 \times 5^2\). Then, since \(N\) is divisible by \(2025\), \(N\) must have at least four factors of \(3\) and at least two factors of \(5\).
\(N\) already has two factors of \(3\) and two factors of \(5\). Thus, \(N\) needs at least two more factors of \(3\) in order to make it divisible by \(2025\). Therefore, the smallest possible value for \(k\) is \(3^2=9\).