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 2)
Line 10: Line 10:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
Consider \${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>.
  
−
We could have </math>\{4,12\}<math>. Similarly, </math>16<math> must be with </math>3<math>, and no matter to which set </math>48<math> is placed into, we will be forced to have </math>a \cdot b =c<math>.
+
We could have <math>\{4,12\}</math>. Similarly, <math>16</math> must be with <math>3</math>, and no matter to which set <math>48</math> is placed into, we will be forced to have <math>a \cdot b =c</math>.
  
−
Now we have </math>\{3,4\}<math>. </math>9<math> must be with </math>12<math>. Then </math>81<math> must be with </math>\{3,4\}<math>. Since </math>27<math> can't be placed in the same set as </math>\{3,4,81\}<math>, </math>27<math> must go with </math>\{9,12\}<math>. But then no matter where </math>243<math> is placed we will have </math>a\cdot b =c<math>.
+
Now we have <math>\{3,4\}</math>. <math>9</math> must be with <math>12</math>. Then <math>81</math> must be with <math>\{3,4\}</math>. Since <math>27</math> can't be placed in the same set as <math>\{3,4,81\}</math>, <math>27</math> must go with <math>\{9,12\}</math>. But then no matter where <math>243</math> is placed we will have <math>a\cdot b =c</math>.
  
−
Thus, </math>\boxed{243}<math> is the minimum </math>m$.
+
Thus, <math>\boxed{243}</math> is the minimum <math>m</math>.
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 21:09, 19 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 $m = \boxed{243}$.

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 $m$.

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=237319"