Difference between revisions of "2002 AIME I Problems/Problem 14"
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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>. | 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 | + | 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</math> <math>\Rightarrow{\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. |
== See also == | == See also == |
Revision as of 06:09, 9 August 2019
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 greatest number of elements that
can have?
Solution
Let the sum of the integers in be
, and let the size of
be
. After any element
is removed, we are given that
, so
. Since
,
, and all elements are congruent to 1 mod
. Since they are positive integers, the largest element is at least
, the
th positive integer congruent to 1 mod
.
We are also given that this largest member is 2002, so , and
. Also, we have
, so
. The largest factor of 2001 less than 45 is 29, so
and
is the largest possible. This can be achieved with
, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.