2024 AMC 10A Problems/Problem 20

Revision as of 20:46, 9 November 2024 by Ihategeometry (talk | contribs) (Solution 1.1)

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 1.1(Faster calculation)==

After finding out that each loop adds 10 each time and has 3 elements, we do a rough calculation by dividing 2024 by 10 and multiplying it by 3, giving us 607.2. Since 608 is the only answer close to that, we figure out that the answer is$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(C) }608}$.$-iHateGeometry

Video Solution by Pi Academy

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


See also

2024 AMC 10A (ProblemsAnswer KeyResources)
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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