The prime factorization of 2007 is $3^{2}\times223$. How many ordered pairs of positive integers $(x,y)$ satisfy the equation $xy=2007$?
$x$ must be a divisor of 2007. To count the number of divisors, we note that a divisor can have zero factors of 3, one factor, or two factors, and (independently) it can have zero factors of 223 or one factor. The total number of divisors is then $3\cdot2=6$ (3 choices for the factors of 3 and 2 choices for the factors of 223). Since each divisor is unique, each creates a unique ordered pair. So there are $\boxed{6}$ ordered pairs.
6
Level 4
Number Theory
Problem: The prime factorization of 2007 is $3^{2}\times223$. How many ordered pairs of positive integers $(x,y)$ satisfy the equation $xy=2007$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the number of ordered pairs of positive integers \((x,y)\) such that \(xy=2007\), we start by recognizing that each ordered pair corresponds to a unique factorization of 2007 into two factors. 

First, we note the prime factorization of 2007, which is \(3^{2} \times 223\). The total number of positive divisors of a number can be calculated using its prime factorization. For a number of the form \(p_1^{e_1} \times p_2^{e_2} \times \ldots \times p_n^{e_n}\), the number of positive divisors \(d\) is given by the formula:

\[ d = (e_1 + 1)(e_2 + 1) \ldots (e_n + 1) \]

Applying this to our factorization:

- For the prime \(3\) with exponent \(2\), we have \(e_1 + 1 = 2 + 1 = 3\).
- For the prime \(223\) with exponent \(1\), we have \(e_2 + 1 = 1 + 1 = 2\).

Thus, the total number of positive divisors of 2007 is:

\[ d = 3 \times 2 = 6 \]

Each divisor \(d_i\) of 2007 can be paired with another divisor \(d_j\) such that \(d_i \times d_j = 2007\). Since the order matters in the pairs \((x,y)\), each divisor can be arranged in two ways: \((d_i, d_j)\) and \((d_j, d_i)\). 

However, since \(d_i\) and \(d_j\) are distinct for all pairs except when \(d_i = d_j\), we can conclude that each of the 6 divisors will contribute to a unique ordered pair. 

Therefore, the total number of ordered pairs \((x,y)\) that satisfy \(xy=2007\) is equal to the number of positive divisors, which is \(6\). 

The final answer is \(6\).<|END_OF_LATENT|> The final answer is $6$.