AMC12 2010 B

AMC12 2010 B · Q14

AMC12 2010 B · Q14. It mainly tests Linear inequalities, Optimization (basic).

Let $a$, $b$, $c$, $d$, and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$, $b+c$, $c+d$ and $d+e$. What is the smallest possible value of $M$?

设 $a$, $b$, $c$, $d$, 和 $e$ 是正整数，且 $a+b+c+d+e=2010$，令 $M$ 为 $a+b$, $b+c$, $c+d$ 和 $d+e$ 这四个和中的最大值。$M$ 的最小可能值是多少？

(A) 670 670

(B) 671 671

(D) 803 803

(E) 804 804

Answer

Correct choice: (B)

正确答案：（B）

Solution

We want to try make $a+b$, $b+c$, $c+d$, and $d+e$ as close as possible so that $M$, the maximum of these, is smallest. Notice that $2010=670+670+670$. In order to express $2010$ as a sum of $5$ numbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as close as possible): $2010=670+1+670+1+668$ or $2010=670+1+669+1+669$. We see that in both cases, the value of $M$ is $671$, so the answer is $671 \Rightarrow \boxed{B}$.

我们希望让 $a+b$, $b+c$, $c+d$ 和 $d+e$ 尽可能接近，从而使它们的最大值 $M$ 尽可能小。注意到 $2010=670+670+670$。为了将 $2010$ 表示为 $5$ 个数之和，我们必须把其中一些数拆分。有两种方式可以做到这一点（同时使两数之和尽可能接近）：$2010=670+1+670+1+668$ 或 $2010=670+1+669+1+669$。可以看到在两种情况下，$M$ 的值都是 $671$，所以答案是 $671 \Rightarrow \boxed{B}$。

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