Difference between revisions of "2010 AIME I Problems/Problem 12"
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For <math>m \le 242</math>, we can partition <math>S</math> into <math>S \cap \{3, 4, 5, 6, 7, 8, 81, 82, 83, 84 ... 242\}</math> and <math>S \cap \{9, 10, 11 ... 80\}</math>, and in neither set are there values where <math>ab=c</math> (since <math>8 < (3\text{ to }8)^2 < 81</math> and <math>81^2>242</math> and <math>(9\text{ to }80)^2 > 80 </math>). Thus <math>m = \boxed{243}</math>. | For <math>m \le 242</math>, we can partition <math>S</math> into <math>S \cap \{3, 4, 5, 6, 7, 8, 81, 82, 83, 84 ... 242\}</math> and <math>S \cap \{9, 10, 11 ... 80\}</math>, and in neither set are there values where <math>ab=c</math> (since <math>8 < (3\text{ to }8)^2 < 81</math> and <math>81^2>242</math> and <math>(9\text{ to }80)^2 > 80 </math>). Thus <math>m = \boxed{243}</math>. | ||
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== See Also == | == See Also == |
Revision as of 00:06, 19 October 2024
Problem
Let be an integer and let
. Find the smallest value of
such that for every partition of
into two subsets, at least one of the subsets contains integers
,
, and
(not necessarily distinct) such that
.
Note: a partition of is a pair of sets
,
such that
,
.
Solution
We claim that is the minimal value of
. Let the two partitioned sets be
and
; we will try to partition
and
such that the
condition is not satisfied. Without loss of generality, we place
in
. Then
must be placed in
, so
must be placed in
, and
must be placed in
. Then
cannot be placed in any set, so we know
is less than or equal to
.
For , we can partition
into
and
, and in neither set are there values where
(since
and
and
). Thus
.
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.