problem: For each positive integer $k$, let $S_k$ denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is $k$.  For example, $S_3$ is the sequence $1,4,7,\ldots$.  For how many values of $k$ does $S_k$ contain $2005$ as a term?
solution: The general term of the sequence is $a_n = 1 + kn$, where $a_0 = 1$ is the first term. Therefore, we want $1 + kn = 2005$, or $kn = 2004$. We see that this equation has a solution for $n$ if and only if $k$ is a divisor of $2004$. Since $2004 = 2^2 \cdot 3 \cdot 167$, the number of positive divisors of $2004$ is $(2+1)(1+1)(1+1) = \boxed{12}$.