For each [positive integer](https://artofproblemsolving.com/wiki/index.php/Positive_integer) $k$, let $S_k$ denote the [ increasing](https://artofproblemsolving.com/wiki/index.php/Increasing_sequence) [arithmetic sequence](https://artofproblemsolving.com/wiki/index.php/Arithmetic_sequence) of [integers](https://artofproblemsolving.com/wiki/index.php/Integer) whose first term is $1$ and whose common difference is $k$.  For example, $S_3$ is the [sequence](https://artofproblemsolving.com/wiki/index.php/Sequence) $1,4,7,10,\ldots.$ For how many values of $k$ does $S_k$ contain the term $2005$?
Suppose that the $n$th term of the sequence $S_k$ is $2005$. Then $1+(n-1)k=2005$ so $k(n-1)=2004=2^2\cdot 3\cdot 167$.  The [ordered pairs](https://artofproblemsolving.com/wiki/index.php/Ordered_pair) $(k,n-1)$ of positive integers that satisfy the last equation are $(1,2004)$,$(2,1002)$, $(3,668)$, $(4,501)$, $(6,334)$, $(12,167)$, $(167,12)$,$(334,6)$, $(501,4)$, $(668,3)$, $(1002,2)$ and $(2004,1)$, and each of these gives a possible value of $k$. Thus the requested number of values is $12$, and the answer is $\boxed{012}$.
Alternatively, notice that the formula for the number of [divisors](https://artofproblemsolving.com/wiki/index.php/Divisor) states that there are $(2 + 1)(1 + 1)(1 + 1) = 12$ divisors of $2^2\cdot 3^1\cdot 167^1$.