The prime factorization of 2007 is $3^{2}\times223$. How many ordered pairs of positive integers $(x,y)$ satisfy the equation $xy=2007$?
I know that the number of ordered pairs of positive integers that satisfy $xy=n$ is equal to the number of positive divisors of $n$. To find the number of positive divisors of $n$, I can use the formula $d(n)=(a+1)(b+1)\cdots(c+1)$, where $a,b,\cdots,c$ are the exponents of the prime factors of $n$. In this case, $n=2007=3^{2}\times223^{1}$, so $a=2$ and $b=1$. Therefore, $d(2007)=(2+1)(1+1)=6$. This means there are 6 ordered pairs of positive integers that satisfy $xy=2007$. 