Difference between revisions of "2005 AIME I Problems/Problem 2"

Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
 
+
Suppose that the ''n''th term of the sequence <math>S_k</math> is 2005. Then <math>1+(n-1)k=2005</math> so <math>k(n-1)=2004=2^2\cdot 3\cdot 167.</math> The ordered pairs <math>(k,n-1)</math> of positive integers that satisfy the last equation are <math>(1,2004)</math>,<math>(2,1002</math>, <math>(3,668)</math>, <math>(4,501)</math>, <math>(6,334)</math>, <math>(12,167)</math>, <math>(167,12)</math>,<math>(334,6)</math>, <math>(501,4)</math>, <math>(668,3)</math>, <math>(1002,2)</math>, and <math>(2004,1)</math>. Thus the requested number of values is 12.
{{solution}}
 
  
 
== See also ==
 
== See also ==
 
* [[2005 AIME I Problems]]
 
* [[2005 AIME I Problems]]

Revision as of 21:46, 28 November 2006

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 nth 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 $(2004,1)$. Thus the requested number of values is 12.

See also

Invalid username
Login to AoPS