[Number Theory]
Problem: Determine the number of positive divisors of 900, including 1 and 900, that are perfect squares. (A positive divisor of 900 is a positive integer that divides exactly into 900.)

Solution: Since $900=30^{2}$ and $30=2 \times 3 \times 5$, then $900=2^{2} 3^{2} 5^{2}$.

The positive divisors of 900 are those integers of the form $d=2^{a} 3^{b} 5^{c}$, where each of $a, b, c$ is 0,1 or 2 .

For $d$ to be a perfect square, the exponent on each prime factor in the prime factorization of $d$ must be even.

Thus, for $d$ to be a perfect square, each of $a, b, c$ must be 0 or 2 .

There are two possibilities for each of $a, b, c$ so $2 \times 2 \times 2=8$ possibilities for $d$.

These are $2^{0} 3^{0} 5^{0}=1,2^{2} 3^{0} 5^{0}=4,2^{0} 3^{2} 5^{0}=9,2^{0} 3^{0} 5^{2}=25,2^{2} 3^{2} 5^{0}=36,2^{2} 3^{0} 5^{2}=100$, $2^{0} 3^{2} 5^{2}=225$, and $2^{2} 3^{2} 5^{2}=900$.

Thus, 8 of the positive divisors of 900 are perfect squares.
The positive divisors of 900 are

$1,2,3,4,5,6,9,10,12,15,18,20,25,30,36,45,50,60,75,90,100,150,180,225,300,450,900$

Of these, $1,4,9,25,36,100,225$, and 900 are perfect squares $\left(1^{2}, 2^{2}, 3^{2}, 5^{2}, 6^{2}, 10^{2}, 15^{2}, 30^{2}\right.$, respectively).

Thus, 8 of the positive divisors of 900 are perfect squares.

Final answer: 8