2022 AMC 10B Problems/Problem 14: Difference between revisions

Latest revision as of 15:56, 9 November 2025

Problem

Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \ldots , 25 \right\}$ 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?

$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$

Solution 1 (Pigeonhole Principle)

Let $M$ be the largest number in $S$ . We categorize numbers $\left\{ 1, 2, \ldots , M-1 \right\}$ (except $\frac{M}{2}$ if $M$ is even) into $\left\lfloor \frac{M-1}{2} \right\rfloor$ 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 $\left\lfloor \frac{M-1}{2} \right\rfloor$ 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 $S = \left\{ 13, 14, \ldots , 25 \right\}$ . 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)

Solution 2 (Pigeonhole v2)

We construct a possible subset $S$ with $13$ elements by including all odd integers from $1$ to $25$ , inclusive. $S=\left\{ 1, 3, 5, \cdots , 25 \right\}$ . The sum of any $2$ elements is even, and thus cannot be an element of $S$ .

To show that $S$ cannot have more than $13$ elements, assume for sake of contradiction that $|S| \geq 14$ . Let $S=\left\{ x_1, x_2, \cdots , x_n \right\}$ where $n \geq 14$ and $x_1 < x_2 < \cdots < x_n$ . Because the sums of any $2$ (not necessarily distinct) elements do not appear in $S$ , $x_1+x_i$ is not an element of $S$ for all $1 \leq i \leq n$ . So, $x_1, x_2, \cdots , x_n , x_1+x_1, x_1+x_2, \cdots , x_1+x_n$ are all distinct integers. Let these integers be elements of the set $T$ . $|T|=2n$ , and because $n \geq 14$ , $|T| \geq 28$ . But all elements of $T$ must be $\geq x_1$ and $\leq x_1+x_n \leq x_1+25$ , leaving only 26 possible values for the elements in $T$ . By the Pigeonhole Principle, the elements cannot be distinct, and we have a contradiction.

Thus, $\boxed{\textbf{(B) }13}$ is the maximum possible size of $S$ .

~starwars101

Solution 3 (Intuitive and Fast)

Notice how $ 24 + 25 > 25 $, $ 23 + 24 > 25 $, and this continues backward until $ n + n+1 > 25 $. We then see that $ 2n+1 > 25 \rightarrow 2n > 24 \rightarrow n > 12 $.

Thus the elements in the set must have the property $ n > 12 : n \in \mathbb{Z}^+ $ and therefore our set is $ \{13, 14, 15, \dots, 25\} $ in which there is $ 25 - 13 + 1 \Rightarrow $ $\boxed{\textbf{(B) }13}$ elements.

~Pinotation

Video Solution

https://youtu.be/_K29sOequlY

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

Video Solution by Interstigation

https://youtu.be/in3N3Os5kGw

Video Solution by TheBeautyofMath

https://youtu.be/Mi2AxPhnRno?t=1056

~IceMatrix

@@ Line 1: / Line 1: @@
 ==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 <math>S</math> is never an element of <math>S.</math> What is the maximum number of elements <math>S</math> may contain?
+Suppose that <imath>S</imath> is a subset of <imath>\left\{ 1, 2, 3, \ldots , 25 \right\}</imath> such that the sum of any two (not necessarily distinct) elements of <imath>S</imath> is never an element of <imath>S.</imath> What is the maximum number of elements <imath>S</imath> may contain?
-<math>\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16</math>
+<imath>\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16</imath>
 ==Solution 1 (Pigeonhole Principle)==
-Let <math>M</math> be the largest number in <math>S</math>.
+Let <imath>M</imath> be the largest number in <imath>S</imath>.
-We categorize numbers <math>\left\{ 1, 2, \ldots , M-1 \right\}</math> (except <math>\frac{M}{2}</math> if <math>M</math> is even) into <math>\left\lfloor \frac{M-1}{2} \right\rfloor</math> groups, such that the <math>i</math>th group contains two numbers <math>i</math> and <math>M-i</math>.
+We categorize numbers <imath>\left\{ 1, 2, \ldots , M-1 \right\}</imath> (except <imath>\frac{M}{2}</imath> if <imath>M</imath> is even) into <imath>\left\lfloor \frac{M-1}{2} \right\rfloor</imath> groups, such that the <imath>i</imath>th group contains two numbers <imath>i</imath> and <imath>M-i</imath>.
-Recall that <math>M \in S</math> and the sum of two numbers in <math>S</math> cannot be equal to <math>M</math>, and the sum of numbers in each group above is equal to <math>S</math>. Thus, each of the above <math>\left\lfloor \frac{M-1}{2} \right\rfloor</math> groups can have at most one number in <math>S</math>.
+Recall that <imath>M \in S</imath> and the sum of two numbers in <imath>S</imath> cannot be equal to <imath>M</imath>, and the sum of numbers in each group above is equal to <imath>S</imath>. Thus, each of the above <imath>\left\lfloor \frac{M-1}{2} \right\rfloor</imath> groups can have at most one number in <imath>S</imath>.
 Therefore,
 <cmath>
@@ Line 19: / Line 19: @@
 </cmath>
-Next, we construct an instance of <math>S</math> with <math>|S| = 13</math>.
+Next, we construct an instance of <imath>S</imath> with <imath>|S| = 13</imath>.
-Let <math>S = \left\{ 13, 14, \ldots , 25 \right\}</math>.
+Let <imath>S = \left\{ 13, 14, \ldots , 25 \right\}</imath>.
 Thus, this set is feasible.
-Therefore, the most number of elements in <math>S</math> is
+Therefore, the most number of elements in <imath>S</imath> is
-<math>\boxed{\textbf{(B) }13}</math>.
+<imath>\boxed{\textbf{(B) }13}</imath>.
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
-==Solution 2==
+== Solution 2 (Pigeonhole v2) ==
-We know that two odd numbers sum to an even number, so we can easily say that odd numbers <math>1-25</math> can be included in the list, making for <math>13</math> elements. But, how do we know we can't include even numbers for a higher element value? Well, to get a higher element value than <math>13</math>, odd numbers as well as even numbers would have to be included in the list (since there are only <math>12</math> even numbers from <math>1-25</math>, and many of those even numbers are the sum of even numbers). However, for every even value we add to our odd list, we have to take away an odd number because there are either two odd numbers that sum to that even value, or that even value and another odd number will sum to an odd number later in the list. So, <math>\boxed{\textbf{(B) }13}</math> elements is the highest we can go.
+We construct a possible subset <imath>S</imath> with <imath>13</imath> elements by including all odd integers from <imath>1</imath> to <imath>25</imath>, inclusive. <imath>S=\left\{ 1, 3, 5, \cdots , 25 \right\}</imath>. The sum of any <imath>2</imath> elements is even, and thus cannot be an element of <imath>S</imath>.
+To show that <imath>S</imath> cannot have more than <imath>13</imath> elements, assume for sake of contradiction that <imath>|S| \geq 14</imath>. Let <imath>S=\left\{ x_1, x_2, \cdots , x_n \right\}</imath> where <imath>n \geq 14</imath> and <imath>x_1 < x_2 < \cdots < x_n</imath>. Because the sums of any <imath>2</imath> (not necessarily distinct) elements do not appear in <imath>S</imath>, <imath>x_1+x_i</imath> is not an element of <imath>S</imath> for all <imath>1 \leq i \leq n</imath>. So, <imath>x_1, x_2, \cdots , x_n , x_1+x_1, x_1+x_2, \cdots , x_1+x_n</imath> are all distinct integers. Let these integers be elements of the set <imath>T</imath>. <imath>|T|=2n</imath>, and because <imath>n \geq 14</imath>, <imath>|T| \geq 28</imath>. But all elements of <imath>T</imath> must be <imath>\geq x_1</imath> and <imath>\leq x_1+x_n \leq x_1+25</imath>, leaving only 26 possible values for the elements in <imath>T</imath>. By the Pigeonhole Principle, the elements cannot be distinct, and we have a contradiction.
+Thus, <imath> \boxed{\textbf{(B) }13}</imath> is the maximum possible size of <imath>S</imath>.
+~starwars101
+==Solution 3 (Intuitive and Fast)==
+Notice how \( 24 + 25 > 25 \), \( 23 + 24 > 25 \), and this continues backward until \( n + n+1 > 25 \). We then see that \( 2n+1 > 25 \rightarrow 2n > 24 \rightarrow n > 12 \).
+Thus the elements in the set must have the property \( n > 12 : n \in \mathbb{Z}^+ \) and therefore our set is \( \{13, 14, 15, \dots, 25\} \) in which there is \( 25 - 13 + 1 \Rightarrow \) <imath> \boxed{\textbf{(B) }13}</imath> elements.
+~Pinotation
 ==Video Solution==
@@ Line 38: / Line 53: @@
 https://youtu.be/in3N3Os5kGw
+==Video Solution by TheBeautyofMath==
+https://youtu.be/Mi2AxPhnRno?t=1056
+~IceMatrix
 == See Also ==
+[https://artofproblemsolving.com/wiki/index.php?title=2025_AMC_12A_Problems/Problem_15 2025 AMC 10A Q21/AMC 12A Q15]
 {{AMC10 box|year=2022|ab=B|num-b=13|num-a=15}}
 {{MAA Notice}}

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