Difference between revisions of "2002 AIME I Problems/Problem 14"
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== Problem == | == Problem == | ||
− | A set <math>\mathcal{S}</math> of distinct positive integers has the following property: for every integer <math>x</math> in <math>\mathcal{S},</math> the arithmetic mean of the set of values obtained by deleting <math>x</math> from <math>\mathcal{S}</math> is an integer. Given that 1 belongs to <math>\mathcal{S}</math> and that 2002 is the largest element of <math>\mathcal{S},</math> what is the | + | A set <math>\mathcal{S}</math> of distinct positive integers has the following property: for every integer <math>x</math> in <math>\mathcal{S},</math> the arithmetic mean of the set of values obtained by deleting <math>x</math> from <math>\mathcal{S}</math> is an integer. Given that 1 belongs to <math>\mathcal{S}</math> and that 2002 is the largest element of <math>\mathcal{S},</math> what is the greatest number of elements that <math>\mathcal{S}</math> can have? |
== Solution == | == Solution == |
Revision as of 20:22, 18 July 2016
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.