Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AIME I Problems/Problem 12"

(→‎Solution 2)
(→‎Solution 1)
 
(2 intermediate revisions by one other user not shown)
Line 8: Line 8:
  
 
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>.
 +
 +
FYI, this. is a bad solution
  
 
== Solution 2 ==
 
== Solution 2 ==
−
Consider <math>\{3,4,12\}</math>. We could have any two of the three be together in the same set, and the third in the other set. Thus, we have <math>{3,4}, {3,12}, {4,12}</math>. We will try to 'place' numbers in either set such that we never have <math>a\cdot b = c</math>, until we reach a point where we MUST have <math>a\cdot b =c</math>.  
+
Consider <math>\{3,4,12\}</math>. We could have any two of the three be together in the same set, and the third in the other set. Thus, we have <math>\{3,4\}, \{3,12\}, \{4,12\}</math>. We will try to 'place' numbers in either set such that we never have <math>a\cdot b = c</math>, until we reach a point where we MUST have <math>a\cdot b =c</math>.  
  
 
We begin with <math>\{3,12\}</math>. Notice that <math>a,b,c</math> do not have to be distinct, meaning we could have <math>3\cdot 3=9</math>. Thus <math>9</math> must be with <math>4</math>. Notice that no matter in which set <math>36</math> is placed, we will be forced to have <math>a\cdot b =c</math>, since <math>3*12=36</math> and <math>4*9=36</math>.
 
We begin with <math>\{3,12\}</math>. Notice that <math>a,b,c</math> do not have to be distinct, meaning we could have <math>3\cdot 3=9</math>. Thus <math>9</math> must be with <math>4</math>. Notice that no matter in which set <math>36</math> is placed, we will be forced to have <math>a\cdot b =c</math>, since <math>3*12=36</math> and <math>4*9=36</math>.
Line 19: Line 21:
  
 
Thus, <math>\boxed{243}</math> is the minimum <math>m</math>.
 
Thus, <math>\boxed{243}</math> is the minimum <math>m</math>.
 +
 +
~skibbysiggy
  
 
== Video Solution ==
 
== Video Solution ==

Latest revision as of 15:29, 26 December 2024

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 1

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 $m = \boxed{243}$.

FYI, this. is a bad solution

Solution 2

Consider $\{3,4,12\}$. We could have any two of the three be together in the same set, and the third in the other set. Thus, we have $\{3,4\}, \{3,12\}, \{4,12\}$. We will try to 'place' numbers in either set such that we never have $a\cdot b = c$, until we reach a point where we MUST have $a\cdot b =c$.

We begin with $\{3,12\}$. Notice that $a,b,c$ do not have to be distinct, meaning we could have $3\cdot 3=9$. Thus $9$ must be with $4$. Notice that no matter in which set $36$ is placed, we will be forced to have $a\cdot b =c$, since $3*12=36$ and $4*9=36$.

We could have $\{4,12\}$. Similarly, $16$ must be with $3$, and no matter to which set $48$ is placed into, we will be forced to have $a \cdot b =c$.

Now we have $\{3,4\}$. $9$ must be with $12$. Then $81$ must be with $\{3,4\}$. Since $27$ can't be placed in the same set as $\{3,4,81\}$, $27$ must go with $\{9,12\}$. But then no matter where $243$ is placed we will have $a\cdot b =c$.

Thus, $\boxed{243}$ is the minimum $m$.

~skibbysiggy

Video Solution

2010 AIME I #12

MathProblemSolvingSkills.com


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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_12&oldid=237939"