The prime factorization of \(20\) is \(2^2\times 5\).
The number \(20\) has \(6\) positive divisors. They are:
\(2^05^0=1\), \(2^05^1=5\), \(2^15^0=2\), \(2^15^1=10\), \(2^25^0=4\), \(2^25^1=20\)
Two of the divisors, \(1\) and \(4\), are perfect squares.
How many positive divisors of \(36^3\) are perfect squares?