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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>(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>(9\text{ to }80)^2 > 80 </math>). Thus <math>m = \boxed{243}</math>.
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 +
==Video Solution==
 +
 +
https://youtu.be/WCPxL5dLKCI
 +
 +
~Shreyas S
  
 
== See Also ==
 
== See Also ==

Revision as of 19:21, 28 June 2020

Problem

Let $m \ge 3$ be an integer and let $S = \{3,4,5,\ldots,m\}$. Find the smallest value of $m$ such that for every partition of $S$ into two subsets, at least one of the subsets contains integers $a$, $b$, and $c$ (not necessarily distinct) such that $ab = c$.

Note: a partition of $S$ is a pair of sets $A$, $B$ such that $A \cap B = \emptyset$, $A \cup B = S$.

Solution

We claim that $243$ is the minimal value of $m$. Let the two partitioned sets be $A$ and $B$; we will try to partition $3, 9, 27, 81,$ and $243$ such that the $ab=c$ condition is not satisfied. Without loss of generality, we place $3$ in $A$. Then $9$ must be placed in $B$, so $81$ must be placed in $A$, and $27$ must be placed in $B$. Then $243$ cannot be placed in any set, so we know $m$ is less than or equal to $243$.

For $m \le 242$, we can partition $S$ into $S \cap \{3, 4, 5, 6, 7, 8, 81, 82, 83, 84 ... 242\}$ and $S \cap \{9, 10, 11 ... 80\}$, and in neither set are there values where $ab=c$ (since $8 < (3\text{ to }8)^2 < 81$ and $(9\text{ to }80)^2 > 80$). Thus $m = \boxed{243}$.

Video Solution

https://youtu.be/WCPxL5dLKCI

~Shreyas S

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

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