2005 AIME I Problems/Problem 2

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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 012.

Alternatively, notice that the formula for the number of divisors states that there are $(2 + 1)(1 + 1)(1 + 1) = 12$ divisors of $2^23^1167^1$.

See also

2005 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions
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