Difference between revisions of "2002 AIME I Problems/Problem 14"
m |
(I doubt this is right, so can someone check???) |
||
| Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
| − | {{ | + | 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 11:27, 11 August 2008
Problem
A set 
of distinct positive integers has the following property: for every integer 
in 
the arithmetic mean of the set of values obtained by deleting from is an integer. Given that 1 belongs to and that 2002 is the largest element of what is the greatet number of elements that can have?
Solution
Let the sum of the integers in be 
. 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 
is a multiple of 
. Now 
, 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 
, and there aren't enough numbers like that. So 
is the maximum.
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 | ||
