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

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== Solution ==
 
== Solution ==
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Let the sum of the integers in <math>\mathcal{S}</math> be <math>N</math>, and let the size of <math>|\mathcal{S}|</math> be <math>n+1</math>. After any element <math>x</math> is removed, we are given that <math>n|N-x</math>, so <math>x\equiv N\pmod{n}</math>. Since <math>x\in\mathcal{S}</math>, <math>N\equiv1\pmod{n}</math>, and all elements are congruent to 1 mod <math>n</math>. Since they are positive integers, the largest element is at least <math>n^2+1</math>, the <math>(n+1)</math>th positive integer congruent to 1 mod <math>n</math>.
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Let the sum of the integers in <math>\mathcal{S}</math> be <math>N</math>, and let the size of <math>|\mathcal{S}|</math> be <math>n+1</math>. After any element <math>x</math> is removed, we are given that <math>n|N-x</math>, so <math>x\equiv N\pmod{n}</math>. Since <math>1\in\mathcal{S}</math>, <math>N\equiv1\pmod{n}</math>, and all elements are congruent to 1 mod <math>n</math>. Since they are positive integers, the largest element is at least <math>n^2+1</math>, the <math>(n+1)</math>th positive integer congruent to 1 mod <math>n</math>.
  
 
We are also given that this largest member is 2002, so <math>2002\equiv1\pmod{n}</math>, and <math>n|2001=3\cdot23\cdot29</math>. Also, we have <math>n^2+1\le2002</math>, so <math>n<45</math>. The largest factor of 2001 less than 45 is 29, so <math>n=29</math> and <math>n+1=\fbox{30}</math> is the largest possible. This can be achieved with <math>\mathcal{S}=\{1,30,59,88,\ldots,813,2002\}</math>, for instance.
 
We are also given that this largest member is 2002, so <math>2002\equiv1\pmod{n}</math>, and <math>n|2001=3\cdot23\cdot29</math>. Also, we have <math>n^2+1\le2002</math>, so <math>n<45</math>. The largest factor of 2001 less than 45 is 29, so <math>n=29</math> and <math>n+1=\fbox{30}</math> is the largest possible. This can be achieved with <math>\mathcal{S}=\{1,30,59,88,\ldots,813,2002\}</math>, for instance.

Revision as of 08:55, 12 February 2010

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 $N$, and let the size of $|\mathcal{S}|$ be $n+1$. After any element $x$ is removed, we are given that $n|N-x$, so $x\equiv N\pmod{n}$. Since $1\in\mathcal{S}$, $N\equiv1\pmod{n}$, and all elements are congruent to 1 mod $n$. Since they are positive integers, the largest element is at least $n^2+1$, the $(n+1)$th positive integer congruent to 1 mod $n$.

We are also given that this largest member is 2002, so $2002\equiv1\pmod{n}$, and $n|2001=3\cdot23\cdot29$. Also, we have $n^2+1\le2002$, so $n<45$. The largest factor of 2001 less than 45 is 29, so $n=29$ and $n+1=\fbox{30}$ is the largest possible. This can be achieved with $\mathcal{S}=\{1,30,59,88,\ldots,813,2002\}$, for instance.

See also

2002 AIME I (Problems • Answer Key • Resources)
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
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