Difference between revisions of "2002 AIME I Problems/Problem 14"

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(I doubt this is right, so can someone check???)
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== Solution ==
 
== Solution ==
{{solution}}
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Let the sum of the integers in <math>\mathcal{S}</math> be <math>S</math>. We are given that <math>\dfrac{S-1}{#(\mathcal{S})-1}</math> and <math>\dfrac{S-2002}{#(\mathcal{S})-1}</math> are integers. Thus <math>2001</math> is a multiple of <math>\mathcal{S}-1</math>. Now <math>2001=3*667</math>, so either <math>#(\mathcal{S})</math> is 2002, 668, 4, or 2. 2 is guaranteed possible, 2002 is not. 4 is: 1, 4, 7, 2002. For 668, all 668 numbers must be congruent mod <math>667</math>, and there aren't enough numbers like that. So <math>004</math> is the maximum.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=I|num-b=13|num-a=15}}
 
{{AIME box|year=2002|n=I|num-b=13|num-a=15}}

Revision as of 12:27, 11 August 2008

Problem

A set $\mathcal{S}$ of distinct positive integers has the following property: for every integer $x$ in $\mathcal{S},$ the arithmetic mean of the set of values obtained by deleting $x$ from $\mathcal{S}$ is an integer. Given that 1 belongs to $\mathcal{S}$ and that 2002 is the largest element of $\mathcal{S},$ what is the greatet number of elements that $\mathcal{S}$ can have?

Solution

Let the sum of the integers in $\mathcal{S}$ be $S$. We are given that $\dfrac{S-1}{#(\mathcal{S})-1}$ (Error compiling LaTeX. Unknown error_msg) and $\dfrac{S-2002}{#(\mathcal{S})-1}$ (Error compiling LaTeX. Unknown error_msg) are integers. Thus $2001$ is a multiple of $\mathcal{S}-1$. Now $2001=3*667$, so either $#(\mathcal{S})$ (Error compiling LaTeX. Unknown error_msg) is 2002, 668, 4, or 2. 2 is guaranteed possible, 2002 is not. 4 is: 1, 4, 7, 2002. For 668, all 668 numbers must be congruent mod $667$, and there aren't enough numbers like that. So $004$ is the maximum.

See also

2002 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions