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 nth 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.