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

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== 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. | |

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== 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 let denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is For example, is the sequence For how many values of does contain the term 2005?

## Solution

Suppose that the *n*th term of the sequence is 2005. Then so The ordered pairs of positive integers that satisfy the last equation are ,, , , , , ,, , , , and . Thus the requested number of values is 12.