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Difference between revisions of "2022 AMC 10B Problems/Problem 14"

(Created page with "==Problem== Suppose that <math>S</math> is a subset of <math>\left\{ 1, 2, 3, \cdots , 25 \right\}</math> such that the sum of any two (not necessarily distinct) elements of...")
 
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Thus, this set is feasible.
 
Thus, this set is feasible.
 
Therefore, the most number of elements in <math>S</math> is
 
Therefore, the most number of elements in <math>S</math> is
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\boxed{\textbf{(B) 13}}.
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<math>\boxed{\textbf{(B) 13}}</math>.
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 16:38, 17 November 2022

Problem

Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S$. What is the maximum number of elements $S$ may contain?

Solution (Pigeonhole Principle)

Denote by $M$ the largest number in $S$. We categorize numbers $\left\{ 1, 2, \cdots , M-1 \right\}$ (except $\frac{M}{2}$ if $M$ is even) into $\lfloor \frac{M-1}{2} \rfloor$ groups, such that the $i$th group contains two numbers $i$ and $M-i$.

Recall that $M \in S$ and the sum of two numbers in $S$ cannot be equal to $M$, and the sum of numbers in each group above is equal to $S$. Thus, each of the above $\lfloor \frac{M-1}{2} \rfloor$ groups can have at most one number in $S$. Therefore, \begin{align*} |S| & \leq 1 + \left\lfloor \frac{M-1}{2} \right\rfloor \\ & \leq 1 + \left\lfloor \frac{25}{2} \right\rfloor \\ & = 13 . \end{align*}

Next, we construct an instance of $S$ with $|S| = 13$. Let $S = \left\{ 13, 14, \cdots , 25 \right\}$. Thus, this set is feasible. Therefore, the most number of elements in $S$ is $\boxed{\textbf{(B) 13}}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/_K29sOequlY

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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