# 2005 AIME I Problems/Problem 2

## 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 $1,4,7,10,\ldots.$ For how many values of $k$ does $S_k$ contain the term $2005$?

## Solution

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 $(k,n-1)$ 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 states that there are $(2 + 1)(1 + 1)(1 + 1) = 12$ divisors of $2^2\cdot 3^1\cdot 167^1$.

## Solution 2

Any term in the sequence $S_k$ can be written as 1+kx. If this is to equal 2005, then the remainder when 2005 is divided by k is 1.

Now all we have to do is find the numbers of factors of 2004. There are $(2 + 1)(1 + 1)(1 + 1) = \boxed{012}$ divisors of $2^2\cdot 3^1\cdot 167^1$.

Note that although the remainder when 2005 divided by 1 is not 1, it still works- $S_1$ would be the sequence of all positive integers, in which 2005 must appear.

~ pi_is_3.14