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

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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>
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==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|before=19|num-a=21}}
 
{{AMC10 box|year=2024|ab=A|before=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:25, 8 November 2024

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 $S$?

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

Solution 1

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 $S.$

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
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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