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Difference between revisions of "2024 AMC 10A Problems/Problem 20"

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<math>\textbf{(A) }436 \qquad \textbf{(B) }506 \qquad \textbf{(C) }608 \qquad \textbf{(D) }654 \qquad \textbf{(E) }675</math>
 
<math>\textbf{(A) }436 \qquad \textbf{(B) }506 \qquad \textbf{(C) }608 \qquad \textbf{(D) }654 \qquad \textbf{(E) }675</math>
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==Video Solution 3 by [[sharmaguy]] (5 mins ⏩)==
 
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https://youtu.be/rFU5EW9VOq8
 
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==Video Solution by Scholars Foundation==
 
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https://www.youtube.com/watch?v=FKOqZau--5w&t=1s
 
  
 
==Solution 1==
 
==Solution 1==
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NOTE:
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'''NOTE:'''
  
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To prove that this is the best we can do, consider adding each element one by one, for the first element, say n. If n is greater than 2, we can choose n - 2 which is always better. Therefore, n = 1 or n = 2.  
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To prove that this is the best we can do, consider adding each element one by one, for the first element, say n. If <math>n>2</math>, we can choose <math>n - 2</math> which is always better. Therefore, <math>n = 1</math> or <math>n = 2.</math>
  
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If n = 2 was optimal, then choose it, then the set of usable numbers in <math>S</math> becomes 5 through 2024. We can transform the usable set of <math>S</math> to <math>Q</math> where <math>Q</math> contains the numbers 1 through 2020. Because we assumed n = 2 was optimal, we can choose n = 2 for the set <math>Q</math> too. Because every element in <math>Q</math> is 4 below the elements of <math>S</math>, choosing 2 in <math>Q</math> would mean choosing 6 in set <math>S</math>. By induction we see that our list would be {2,6,10,14,18,.....2022} which only gives 506 elements which is sub-optimal. Therefore, we can conclude that n = 1 is optimal, and we proceed as the solution above.
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If <math>n = 2</math> was optimal, then choose it, then the set of usable numbers in <math>S</math> becomes 5 through 2024. We can transform the usable set of <math>S</math> to <math>Q</math> where <math>Q</math> contains the numbers 1 through 2020. Because we assumed <math>n = 2</math> was optimal, we can choose <math>n = 2</math> for the set <math>Q</math> too. Because every element in <math>Q</math> is 4 below the elements of <math>S</math>, choosing 2 in <math>Q</math> would mean choosing 6 in set <math>S</math>. By induction we see that our list would be <math>\{2,6,10,14,18,.....2022\}</math> which only gives <math>506</math> elements which is sub-optimal. Therefore, we can conclude that <math>n = 1</math> is optimal, and we proceed as the solution above.
  
 
-weihou0
 
-weihou0
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 +
- minor latex edits by eqb5000
  
 
==Solution 2==
 
==Solution 2==
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==Solution 3==  
 
==Solution 3==  
  
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Experiment with the first 24 numbers and consecutively pair them. We find the following pairs of numbers work:  
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We find the following pairs of numbers work:  
  
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<math>(1,4), (4,8), (8,11), (11,14), (14, 18), (18, 21), (21, 24).</math>
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<cmath>(1,4), (4,8), (8,11), (11,14), (14, 18), (18, 21), (21, 24) \dots</cmath>
  
  
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Call a number an OE (odd-even) pair if the first number in the pair is odd and the second is even. Do the same for EE and EO pairs. Notice that we're adding 10 to the pairs in each set, and there are 404 numbers in each set of pairs excluding 1 and 2024:  
+
Call a number an OE (odd-even) pair if the first number in the pair is odd and the second is even. Do the same for EE (even-even) and EO (even-odd) pairs. Notice that we're adding 10 to the pairs in each set, so there are 404 numbers in each set of pairs '''excluding 1 and 2024:'''
  
  
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'''OE pairs:''' <math>(1,4), (11,14), ... , (2021,2024) \implies (202 + 1) \cdot 2 - 2 = 404</math> numbers
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'''OE pairs:''' <cmath>(1,4), (11,14), ... , (2021,2024) \implies (202 + 1) \cdot 2 - 2 = 404</cmath>  
 
   
 
   
  
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'''EE pairs:''' <math>(4,8), (14,18), ..., (2014,2018)  \implies (201 + 1) \cdot 2 = 404</math> numbers
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'''EE pairs:''' <cmath>(4,8), (14,18), ..., (2014,2018)  \implies (201 + 1) \cdot 2 = 404</cmath>  
  
  
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'''EO pairs:''' <math>(8,11), (18,21), ..., (2018, 2021) \implies (201 + 1) \cdot 2 = 404</math> numbers
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'''EO pairs:''' <cmath>(8,11), (18,21), ..., (2018, 2021) \implies (201 + 1) \cdot 2 = 404</cmath>
  
  
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Every number (besides 1 and 2024 which we'll add back later) is being overcounted twice, so we have <math>404 \cdot 3/2 = 606.</math> Finally, we add 2 to get <math>\boxed{\textbf{(C) }608}</math>.
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Every number besides <math>1</math> and <math>2024</math> is being overcounted twice (for example, <math>4</math> is counted twice in the EE and OE pairs), so we have <math>404 \cdot \frac{3}{2} = 606.</math> Finally, we add <math>2</math> (adding back <math>1</math> and <math>2024</math>) to get <math>\boxed{\textbf{(C) }608}</math>.
  
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~[[sharmaguy]]
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~[https://artofproblemsolving.com/wiki/index.php/User:grogg007 grogg007]
  
 
== Video Solution by Power Solve ==
 
== Video Solution by Power Solve ==
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https://youtu.be/BhiczrT7Sg0?si=XnkHOJl5n9SWHsfc
 
https://youtu.be/BhiczrT7Sg0?si=XnkHOJl5n9SWHsfc
  
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==See also==
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==Video Solution 3 by [https://artofproblemsolving.com/wiki/index.php/User:grogg007 grogg007] (5 mins ⏩)==
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{{AMC10 box|year=2024|ab=A|num-b=19|num-a=21}}
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https://youtu.be/rFU5EW9VOq8
 +
 
 +
==Video Solution by Scholars Foundation==
 +
https://www.youtube.com/watch?v=FKOqZau--5w&t=1s
 +
 
 +
== Video solution by TheNeuralMathAcademy ==
 +
https://www.youtube.com/watch?v=4b_YLnyegtw&t=4268s
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2024|ab=A|before=[[2023 AMC 10B Problems]]|after=[[2024 AMC 10B Problems]]}}
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* [[AMC 10]]
 +
* [[AMC 10 Problems and Solutions]]
 +
* [[Mathematics competitions]]
 +
* [[Mathematics competition resources]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:46, 27 September 2025

Problem

Let $S$ be a subset of $\{1, 2, 3, \dots, 2024\}$ such that the following two conditions hold:

  • If $x$ and $y$ are distinct elements of $S$, then $|x-y| > 2.$
  • If $x$ and $y$ are distinct odd elements of $S$, then $|x-y| > 6.$

What is the maximum possible number of elements in $S$?

$\textbf{(A) }436 \qquad \textbf{(B) }506 \qquad \textbf{(C) }608 \qquad \textbf{(D) }654 \qquad \textbf{(E) }675$

Solution 1

All lists are organized in ascending order:

By listing out the smallest possible elements of subset $S,$ we can find that subset $S$ starts with $\{1, 4, 8, 11, 14, 18, 21, 24, 28, 31, \dots\}.$ It is easily noticed that the elements of the subset "loop around" every 3 elements, specifically adding 10 each time. This means that there will be $2024/10$ or $202R4$ whole loops in the subset $S,$ implying that there will be $202*3 = 606$ elements in S. However, we have undercounted, as we did not count the remainder that resulted from $2024/10$$.$ With a remainder of $4,$ we can fit $2$ more elements into the subset $S,$ namely $2021$ and $2024,$ resulting in a total of $606+2$ or $\boxed{\textbf{(C) }608}$ elements in subset $S$.


NOTE:

To prove that this is the best we can do, consider adding each element one by one, for the first element, say n. If $n>2$, we can choose $n - 2$ which is always better. Therefore, $n = 1$ or $n = 2.$

If $n = 2$ was optimal, then choose it, then the set of usable numbers in $S$ becomes 5 through 2024. We can transform the usable set of $S$ to $Q$ where $Q$ contains the numbers 1 through 2020. Because we assumed $n = 2$ was optimal, we can choose $n = 2$ for the set $Q$ too. Because every element in $Q$ is 4 below the elements of $S$, choosing 2 in $Q$ would mean choosing 6 in set $S$. By induction we see that our list would be $\{2,6,10,14,18,.....2022\}$ which only gives $506$ elements which is sub-optimal. Therefore, we can conclude that $n = 1$ is optimal, and we proceed as the solution above.

-weihou0

- minor latex edits by eqb5000

Solution 2

Notice that we can first place odd numbers, then place even numbers between each pair. We can start at $1$ and continue from there. Realize that the smallest number $k$ such that $kx+1$ reproduces odd number is $8$. The next ones are $10, 12, 14$. We can proceed to find the number of numbers in this particular sequence. From the equation $8x+1=2023$, we get that $x \leq 252.875$ works, so this means there is 253 solutions. Looking at $1,2,3,4,5,6,7,9$ we can see that there could only be 1 possible number between each pair, yielding $252+253=505$. Then see that we can fit two more into the number count since the set $2017$ to $2024$ can fit two evens. Now this means $A$ and $B$ don’t work. Now test out $10x+1$. Using the same method, we get that $608$ is the maximum number in the set. Everything above $x=10$ doesn’t work, as we can split it down into smaller subgroups, so the answer is $\boxed{\textbf{(C) }608}$.

~EaZ_Shadow

Solution 3

We find the following pairs of numbers work:

\[(1,4), (4,8), (8,11), (11,14), (14, 18), (18, 21), (21, 24) \dots\]


Call a number an OE (odd-even) pair if the first number in the pair is odd and the second is even. Do the same for EE (even-even) and EO (even-odd) pairs. Notice that we're adding 10 to the pairs in each set, so there are 404 numbers in each set of pairs excluding 1 and 2024:


OE pairs: \[(1,4), (11,14), ... , (2021,2024) \implies (202 + 1) \cdot 2 - 2 = 404\]


EE pairs: \[(4,8), (14,18), ..., (2014,2018) \implies (201 + 1) \cdot 2 = 404\]


EO pairs: \[(8,11), (18,21), ..., (2018, 2021) \implies (201 + 1) \cdot 2 = 404\]


Every number besides $1$ and $2024$ is being overcounted twice (for example, $4$ is counted twice in the EE and OE pairs), so we have $404 \cdot \frac{3}{2} = 606.$ Finally, we add $2$ (adding back $1$ and $2024$) to get $\boxed{\textbf{(C) }608}$.

~grogg007

Video Solution by Power Solve

https://www.youtube.com/watch?v=NZ0SBMqeAfg

Video Solution by Pi Academy

https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE

Video Solution by SpreadTheMathLove

https://youtu.be/BhiczrT7Sg0?si=XnkHOJl5n9SWHsfc

Video Solution 3 by grogg007 (5 mins ⏩)

https://youtu.be/rFU5EW9VOq8

Video Solution by Scholars Foundation

https://www.youtube.com/watch?v=FKOqZau--5w&t=1s

Video solution by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=4268s

See Also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
2023 AMC 10B Problems 		Followed by
2024 AMC 10B Problems
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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