2010 AMC 12B Problems/Problem 14

Problem 14

Let $a$ , $b$ , $c$ , $d$ , and $e$ be postive 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 $M$ ?

$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$

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, if 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 $671 \Rightarrow \boxed{B}$ .

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2010 AMC 12B Problems/Problem 14

Problem 14

Solution

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