AMC10 2024 A

AMC10 2024 A · Q20

AMC10 2024 A · Q20. It mainly tests Basic counting (rules of product/sum), Casework.

Let $S$ be a subset of $\{1, 2, 3, \dots, 2024\}$ such that the following two conditions hold: What is the maximum possible number of elements in $S$?

设$S$为集合$\{1, 2, 3, \dots, 2024\}$的子集，使得以下两个条件成立： $S$的最大可能元素个数是多少？

(A) 436 436

(B) 506 506

(D) 654 654

(E) 675 675

Answer

Correct choice: (C)

正确答案：（C）

Solution

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 $\boxed{\textbf{(C) }608}$

所有列表按升序排列：通过列出子集$S$的最小可能元素，可得$S$以$\{1, 4, 8, 11, 14, 18, 21, 24, 28, 31, \dots\}$开头。容易注意到子集元素每$3$个“循环”一次，每次加$10$。这意味着$S$中有$2024/10=202$整循环（余$4$），即$202 \times 3 = 606$个元素。但未计入余数部分，余$4$可再容纳$2$个元素，即$2021$和$2024$，总计$606 + 2 = \boxed{\textbf{(C) }608}$。

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