AMC10 2024 B

AMC10 2024 B · Q5

AMC10 2024 B · Q5. It mainly tests Inequalities (AM-GM etc. basic), Arithmetic misc.

In the following expression, Melanie changed some of the plus signs to minus signs: \[1+3+5+7+...+97+99\] 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?

在以下表达式中，Melanie将一些加号改为减号： \[1+3+5+7+...+97+99\] 新表达式计算后为负数。Melanie最少改动了多少个加号为减号？

(A) 14 14

(B) 15 15

(D) 17 17

(E) 18 18

Answer

Correct choice: (B)

正确答案：（B）

Solution

Recall that the sum of the first $n$ odd numbers is $n^2$. Thus \[1 + 3 + 5 + 7+ \dots + 97 + 99 = 50^2 = 2500.\] If we want to minimize the number of sign flips to make the number negative, we must flip the signs corresponding to the values with largest absolute value. This will result in the inequality \[1 + 3 + 5 +\dots + (2n - 3) + (2n - 1) - (2n + 1) - (2n + 3)-\dots - 97 - 99 < 0.\] The positive section of the sum will contribute $n^2$, and the negative section will contribute $-(2500-n^2) = (n^2 - 2500)$. The inequality simplifies to \[n^2 + (n^2 - 2500) < 0\] \[2n^2 < 2500\] \[n^2 < 1250\] The greatest positive value of $n$ satisfying the inequality is $n = 35$, corresponding to $35$ positive numbers, and $\boxed{\text{B. } 15}$ negatives. ALEX

回忆前 $n$ 个奇数和为 $n^2$。因此 \[1 + 3 + 5 + 7+ \dots + 97 + 99 = 50^2 = 2500。\] 要使总和为负且最小化符号翻转次数，应翻转绝对值最大的项对应的符号。这样，正部分贡献 $n^2$，负部分贡献 $-(2500-n^2) = (n^2 - 2500)$。不等式简化为 \[n^2 + (n^2 - 2500) < 0\] \[2n^2 < 2500\] \[n^2 < 1250\] 最大整数 $n=35$，对应35个正数和 $\boxed{\text{B. } 15}$ 个负数。

Topics