Question

在下面的式子中，Melanie 把一些加号改成了减号：

$1 + 3 + 5 + 7 + \cdots + 97 + 99$

新式子计算后的值是负数。她最少要把多少个加号改成减号？

In the following expression, Melanie changed some of the plus signs to minus signs:

$1 + 3 + 5 + 7 + \cdots + 97 + 99$

When the new expression was evaluated, it was negative. What is the least number of plus signs that Melanie could have changed to minus signs?

$14$

$15$

$16$

$17$

$18$

答案：B

知识点：求和最优化极限情形界定

难度评级：1250

小提示：

完整和为 $1 + 3 + \cdots + 99 = 50^2 = 2500$ ；把一项 $t$ 变号会使总和减少 $2t$ 。

The full sum is $1 + 3 + \cdots + 99 = 50^2 = 2500;$ flipping a term $t$ lowers the sum by $2t$

大提示：

若想用最少的变号使总和小于 $0$ ，应优先变号最大的项；这些项的两倍总和必须超过 $2500$ 。

To go below $0$ with the fewest flips, flip the largest terms; their doubled total must exceed $2500$

解答：

完整和为 $1 + 3 + \cdots + 99 = 50^2 = 2500$ 。把一项 $t$ 从加号改为减号，会使总和减少 $2t$ ，所以被变号的项之和必须大于 $1250$ 。为了使用最少项，应选择最大的奇数。最大的 $k$ 个奇数之和为 $99 + 97 + \cdots = k(100 - k)$ 。需要 $k(100 - k) \gt 1250$ 。当 $k = 14$ 时只有 $1204$ ，当 $k = 15$ 时为 $1275$ 。所以 $15$ 次变号可以做到。因此正确答案是 B。

The full sum is $1 + 3 + \cdots + 99 = 50^2 = 2500.$ Flipping a term $t$ drops the total by $2t,$ so to go negative the flipped terms have to add up to more than $1250.$ The greedy move is to flip the biggest odd numbers: flipping the top $k$ gives $99 + 97 + \cdots = k(100 - k).$ We want $k(100 - k) \gt 1250.$ At $k = 14$ it’s only $1204,$ but at $k = 15$ it jumps to $1275.$ So $15$ flips do it. Thus, B is the correct answer.

2024 AMC 10B 第 5 题

答案：B

难度评级：1250

小提示：

大提示：

解答：

其他年份的第 5 题