Question

设 $S$ 为 $\{1, 2, 3, \ldots, 2024\}$ 的一个子集，满足以下两个条件：

• 若 $x$ 和 $y$ 是 $S$ 中不同元素，则 $|x - y| \gt 2$ 。

• 若 $x$ 和 $y$ 是 $S$ 中不同的奇数元素，则 $|x - y| \gt 6$ 。

$S$ 最多可能有多少个元素？

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

• If $x$ and $y$ are distinct elements of $S,$ then $|x - y| \gt 2.$

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

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

$436$

$506$

$608$

$654$

$675$

答案：C

知识点：有限制的排列极端原理最优化

难度评级：2080

解答：

两个条件说明选中数至少相差 $3$ ，而选中的奇数至少相差 $7$ 。构造序列 $r,r+3,r+6,r+9$ ，也就是取模 $10$ 余 $6$ 的数。相邻差都 $2$ ，且每个长度为十的块中只有一个奇数，所以选中的奇数彼此相差 $10$ 。每 $10$ 个数取 $3$ 个。集合 $202\cdot3+2=608$ 有 $1,4,8,11,14,18,\ldots$ 个完整的长度为十的块，再加上 $1,4,8\pmod{10}$ ，所以可取 $2021,2024$ 。每个长度为 $3$ 的块最多容纳 $3$ 个元素，所以不能更多，正确答案是 C。

The two conditions say chosen numbers are at least $3$ apart, and chosen odd numbers at least $7$ apart. Any four numbers in a block of $10$ would need three gaps of at least $3,$ so they would have to occupy positions $r,r+3,r+6,r+9.$ Two of the odd entries would then differ by $6,$ which is forbidden. Thus each full block of $10$ contains at most $3$ choices, and the last four positions contain at most $2.$ This gives the upper bound $202\cdot3+2=608.$ It is attained by the pattern $1,4,8,11,14,18,\ldots$ (residues $1,4,8\pmod{10}$ ), together with $2021,2024.$ Adjacent selected values differ by at least $3,$ and the selected odd values are $10$ apart. Therefore, the answer is C.

2024 AMC 10A 第 20 题

答案：C

难度评级：2080

解答：

其他年份的第 20 题