Difference between revisions of "2002 AIME I Problems/Problem 14"
m (→Solution) |
m (667 factorizes =p) |
||
| 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{ | + | 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</math>. We are given that <math>\dfrac{N-1}{n-1}</math> and <math>\dfrac{N-2002}{n-1}</math> are integers. Thus <math>2001</math> is a multiple of <math>n-1</math>. Now <math>2001= 3 \times 23 \times 29</math>, {{incomplete}} |
== 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}} | ||
| + | |||
| + | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 11:58, 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 
, and let the size of 
be 
. We are given that 
and 
are integers. Thus 
is a multiple of 
. Now 
, Template:Incomplete
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 | ||
