AMC12 2025 B

AMC12 2025 B · Q14

AMC12 2025 B · Q14. It mainly tests Linear equations, Sequences & recursion (algebra).

Consider a decreasing sequence of n positive integers \[x_1 > x_2 > \cdots > x_n\] that satisfies the following conditions: What is the greatest possible value of n?

考虑一个由$n$个正整数组成的降序列 \[x_1 > x_2 > \cdots > x_n\] 满足以下条件： $n$的最大可能值是多少？

(A) 1013 1013

(B) 1014 1014

(D) 2016 2016

(E) 2025 2025

Answer

Correct choice: (B)

正确答案：（B）

Solution

Since the mean of the first three terms is $2025$, their sum is $2025 * 3 = 6075$. Then, incorporating the fourth term makes the mean $2025-1=2024$, so their sum must be $2024 * 4 = 8096$, implying that the 4th term is $8096-6075=2021$ Doing the same for the 5th term, $2023 * 5= 10115$, 5th term is $10115-8096=2019$ Algebraically, we can model this pattern for the $k$th term = x as $(k-1)(2029-k) + x = (k)(2028-k)$ $2029k-2029-k^2+k+x=2028k-k^2$ $x=2029-2k$ So the maximum $k\le n$ for which $x$ is positive is 1014, giving our answer $n=\boxed{\textbf{(B) }1014}$

由于前三个项的平均数是$2025$，它们的和是$2025\times3=6075$。然后，加入第四项使平均数为$2025-1=2024$，所以它们的和必须是$2024\times4=8096$，因此第4项是$8096-6075=2021$。对第5项同样操作，$2023\times5=10115$，第5项是$10115-8096=2019$。代数上，我们可以对第$k$项$x$建模为$(k-1)(2029-k)+x=(k)(2028-k)$ $2029k-2029-k^2+k+x=2028k-k^2$ $x=2029-2k$ 因此$x$为正的最大$k\le n$是1014，给出答案$n=\boxed{\textbf{(B) }1014}$

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