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

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

@@ Line 27: / Line 27: @@
 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.
-<math>-weihou0
+-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{\boxed{\textbf{C}}}$.
+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
-https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
+==Video Solution by SpreadTheMathLove==
+https://www.youtube.com/watch?v=6SQ74nt3ynw
 ==See also==
 {{AMC10 box|year=2024|ab=A|num-b=19|num-a=21}}
 {{MAA Notice}}

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