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

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==Problem==
 
==Problem==
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Let <math>S</math> be a subset of <math>\{1, 2, 3, \dots, 2024\}</math> such that the following two conditions hold:
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Let <math>S</math> be a subset of <math>\{1, 2, 3, \dots, 2024\}</math> such that the following two conditions hold: <math>\linebreak</math>
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- If <math>x</math> and <math>y</math> are distinct elements of <math>S</math>, then <math>|x-y| > 2</math>
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* If <math>x</math> and <math>y</math> are distinct elements of <math>S</math>, then <math>|x-y| > 2</math>  <math>\linebreak</math>
−
- If <math>x</math> and <math>y</math> are distinct odd elements of <math>S</math>, then <math>|x-y| > 6</math>.
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* If <math>x</math> and <math>y</math> are distinct odd elements of <math>S</math>, then <math>|x-y| > 6</math>.   <math>\linebreak</math>
 
What is the maximum possible number of elements in <math>S</math>?
 
What is the maximum possible number of elements in <math>S</math>?
  
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\textbf{(D) }654 \qquad
 
\textbf{(D) }654 \qquad
 
\textbf{(E) }675 \qquad</math>
 
\textbf{(E) }675 \qquad</math>
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=Solution 1=
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−
By listing out the smallest possible elements of subset <math>S,</math> we can find that subset <math>S</math> starts with <math>\{1, 4, 8, 11, 14, 18, 21, 24, 28, 31, \dots\}.</math> 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 <math>2024/10</math> or <math>202R4</math> whole loops in the subset <math>S,</math> implying that there will be <math>202*3 = 606</math> elements in S. However, we have undercounted, as we did not count the remainder that resulted from <math>2024/10</math><math>.</math> With a remainder of <math>4,</math> we can fit <math>2</math> more elements into the subset <math>S,</math> namely <math>2021</math> and <math>2024,</math> resulting in a total of <math>606+2</math> or <math>\boxed{\textbf{(C) }608}</math> elements in subset <math>S.</math>
+
==Video Solution by Scholars Foundation==
 +
https://www.youtube.com/watch?v=FKOqZau--5w&t=1s
 +
 
 +
==Solution 1==
 +
All lists are organized in ascending order:
 +
 
 +
By listing out the smallest possible elements of subset <math>S,</math> we can find that subset <math>S</math> starts with <math>\{1, 4, 8, 11, 14, 18, 21, 24, 28, 31, \dots\}.</math> 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 <math>2024/10</math> or <math>202R4</math> whole loops in the subset <math>S,</math> implying that there will be <math>202*3 = 606</math> elements in S. However, we have undercounted, as we did not count the remainder that resulted from <math>2024/10</math><math>.</math> With a remainder of <math>4,</math> we can fit <math>2</math> more elements into the subset <math>S,</math> namely <math>2021</math> and <math>2024,</math> resulting in a total of <math>606+2</math> or <math>\boxed{\textbf{(C) }608}</math> elements in subset <math>S</math>.
 +
 
 +
 
 +
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 is greater than 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 <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.
 +
 
 +
-weihou0
 +
 
 +
==Solution 2==
 +
Notice that we can first place odd numbers, then place even numbers between each pair. We can start at <math>1</math> and continue from there. Realize that the smallest number <math>k</math> such that <math>kx+1</math> reproduces odd number is <math>8</math>. The next ones are <math>10, 12, 14</math>. We can proceed to find the number of numbers in this particular sequence. From the equation <math>8x+1=2023</math>, we get that <math>x \leq 252.875</math> works, so this means there is 253 solutions. Looking at <math>1,2,3,4,5,6,7,9</math> we can see that there could only be 1 possible number between each pair, yielding <math>252+253=505</math>. Then see that we can fit two more into the number count since the set <math>2017</math> to <math>2024</math> can fit two evens. Now this means <math>A</math> and <math>B</math> don’t work. Now test out <math>10x+1</math>. Using the same method, we get that <math>608</math> is the maximum number in the set. Everything above <math>x=10</math> doesn’t work, as we can split it down into smaller subgroups, so the answer is <math>\boxed{\textbf{(C) }608}</math>.
 +
 
 +
~EaZ_Shadow
 +
 
 +
 
 +
== 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://www.youtube.com/watch?v=6SQ74nt3ynw
  
 
==See also==
 
==See also==
−
{{AMC10 box|year=2024|ab=A|before=19|num-a=21}}
+
{{AMC10 box|year=2024|ab=A|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:06, 17 November 2024

Problem

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

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

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

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

Video Solution by Scholars Foundation

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

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 is greater than 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

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


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://www.youtube.com/watch?v=6SQ74nt3ynw

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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

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