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

小提示：

每次操作把 $4$ 个数变成 $1$ 个数，所以数量减少 $3$ ；要最大化剩余数量，就要最小化操作次数。

Each move turns $4$ numbers into $1,$ so the count drops by $3;$ maximizing the count means minimizing the moves

大提示：

共有 $1012$ 个偶数，而每次操作至多使偶数的个数减少 $3$ 个。

There are $1012$ even entries, and any move decreases their number by at most $3$

解答：

在 $1, \ldots, 2024$ 中有 $1012$ 个偶数和 $1012$ 个奇数。每次操作把 $4$ 个数换成 $1$ 个数，所以总数减少 $3$ 。如果某次操作用掉 $e$ 个偶数，那么它写下的结果要么是奇数，使偶数的个数减少 $e$ ；要么是偶数，使偶数的个数减少 $e-1$ 。无论哪种情形，偶数的个数至多减少 $3$ 。因此要消去全部 $1012$ 个偶数，至少需要 $\lceil\frac{1012}{3}\rceil=338$ 次操作。这个次数可以达到：先做 $337$ 次求和，每次取一个奇数和三个偶数，再做一次求和，取三个奇数和最后一个偶数。这样每次写下的都是奇数。因此白板上最多能剩下 $2024-3\cdot338=1010$ 个数。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 $\lceil\frac{1012}{3}\rceil=338$ 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

小提示：

大提示：

解答：

其他年份的第 16 题