Difference between revisions of "2006 USAMO Problems/Problem 2"
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== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
Let one optimal set of integers be <math>\{a_1,\dots,a_{2k+1}\}</math> with <math>a_1 > a_2 > \cdots > a_{2k+1} > 0</math>. | Let one optimal set of integers be <math>\{a_1,\dots,a_{2k+1}\}</math> with <math>a_1 > a_2 > \cdots > a_{2k+1} > 0</math>. | ||
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Hence the smallest possible <math>N</math> is <math>\boxed{ N = 2k^3 + 3k^2 + 3k }</math>. | Hence the smallest possible <math>N</math> is <math>\boxed{ N = 2k^3 + 3k^2 + 3k }</math>. | ||
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+ | {{alternate solutions}} | ||
== See also == | == See also == |
Revision as of 09:47, 5 August 2014
Contents
Problem
(Dick Gibbs) For a given positive integer find, in terms of
, the minimum value of
for which there is a set of
distinct positive integers that has sum greater than
but every subset of size
has sum at most
.
Solution
Solution 1
Let one optimal set of integers be with
.
The two conditions can now be rewritten as and
.
Subtracting, we get that
, and hence
.
In words, the sum of the
smallest numbers must exceed the sum of the
largest ones.
Let . As all the numbers are distinct integers, we must have
, and also
.
Thus we get that , and
.
As we want the second sum to be larger, clearly we must have .
This simplifies to
.
Hence we get that:
On the other hand, for the set the sum of the largest
elements is exactly
, and the sum of the entire set is
, which is more than twice the sum of the largest set.
Hence the smallest possible is
.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
- <url>viewtopic.php?t=84550 Discussion on AoPS/MathLinks</url>
2006 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.