AIME 2010 I · 第 12 题

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 $m$.

~skibbysiggy

Video Solution

2010 AIME I #12

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