Question

Jerry 喜欢玩数字。有一天，他把从 $1$ 到 $2024$ 的所有整数都写在白板上。然后他反复选择白板上的四个数，擦掉它们，并用它们的和或积替代。（例如，Jerry 的第一步可能擦掉 $1, 2, 3$ 和 $5$ ，然后在白板上写下它们的和 $11$ ，或它们的积 $30$ 。）反复进行这个操作后，Jerry 注意到白板上剩下的所有数都是奇数。此时白板上最多可能还剩多少个整数？

Jerry likes to play with numbers. One day, he wrote all the integers from $1$ to $2024$ on the whiteboard. Then he repeatedly chose four numbers on the whiteboard, erased them, and replaced them by either their sum or their product. (For example, Jerry's first step might have been to erase $1, 2, 3,$ and $5,$ and then write either $11,$ their sum, or $30,$ their product, on the whiteboard.) After repeatedly performing this operation, Jerry noticed that all the remaining numbers on the whiteboard were odd. What is the maximum possible number of integers on the whiteboard at that time?

$1010$

$1011$

$1012$

$1013$

$1014$

答案：A

知识点：奇偶性不变量最优化

难度评级：1800

解答：

在 $1, \ldots, 2024$ 中，有 $1012$ 个偶数和 $1012$ 个奇数。每次操作消耗 $4$ 个数并写回 $1$ 个数，所以总数减少 $3$ ；为了剩得最多，需要操作次数尽量少。所有偶数最终都必须消失，而一次产生奇数结果的操作至多能同时清除 $3$ 个偶数（一个奇数加三个偶数的和为奇数）。因此清除全部 $1012$ 个偶数至少需要 $e$ 次操作。这可以达到：用 $337$ 次“一个奇数 $e$ 三个偶数 $e-1$ 奇数和”消去 $\lceil1012/3\rceil=338$ 个偶数，再用一次“三个奇数 $2024-3\cdot338=1010$ 一个偶数奇数和”消去最后一个偶数。剩余数量为。因此正确答案是 A。

Among $1, \ldots, 2024$ there are $1012$ even numbers and $1012$ odd numbers. Each operation replaces $4$ entries by $1,$ so the total count falls by $3.$ If a move consumes $e$ even entries, its output is either odd, reducing the even count by $e,$ or even, reducing it by $e-1.$ In either case the even count falls by at most $3.$ Therefore eliminating all $1012$ even entries takes at least $\lceil1012/3\rceil=338$ moves. This is achievable: use $337$ sums containing one odd and three evens, then one sum containing three odds and the final even. Every output is odd. Thus the maximum remaining count is $2024-3\cdot338=1010.$ Therefore, the answer is A.

2024 AMC 10B 第 16 题

答案：A

难度评级：1800

解答：

其他年份的第 16 题