2010 AIME I Problems/Problem 12

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 $81^2>242$ and $(9\text{ to }80)^2 > 80$ ). Thus $m = \boxed{243}$ .

Video Solution

2010 AIME I #12

MathProblemSolvingSkills.com

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

Page

Toolbox

Search

2010 AIME I Problems/Problem 12

Contents

Problem

Solution

Video Solution

See Also