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?
First, let’s look at the prime factorization of four different perfect squares:
\(9=3^2\), \(16=2^4\), \(36=2^2\times 3^2\), \(129\,600=2^6\times 3^4\times 5^2\)
Note that, in each case, the exponent on each of the prime factors is even. In fact, a positive integer is a perfect square exactly when the exponent on each of the prime factors in its prime factorization is an even integer greater than or equal to zero. Now \[\begin{aligned} 36^3&=(2^2\times 3^2)^3\\ &=(2^2)^3\times (3^2)^3\\ &=2^6\times 3^6 \end{aligned}\] All positive divisors of \(36^3\) will be of the form \(2^k\times 3^n\) where \(k\) and \(n\) are integers with \(0\leq k \leq 6\) and \(0 \leq n \leq 6\).
For \(2^k\times 3^n\) to be a perfect square, \(k\) and \(n\) must be even integers. Thus, \(k \in \{0,2,4,6\}\) and \(n \in \{0,2,4,6\}\).
For each of the \(4\) values of \(k\), there are \(4\) values of \(n\), so there are \(4\times 4= 16\) perfect square divisors of \(36^3\).
Therefore, \(36^3\) has \(16\) divisors that are perfect squares.
We can systematically list all of the divisors that are perfect squares. They are: