2010 AMC 12B Problems/Problem 14

Problem 14

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 $M$ ?

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

Solution 1

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 $671 \Rightarrow \boxed{B}$ .

Solution 2

First, note that, simply by pigeonhole, at least one of a, b, c, d, e is greater than or equal to $\frac{2010}{5}=402,$ so none of C, D, or E can be the answer. Thus, the answer is A or B. We will show that A is unattainable, leaving us with B as the only possible answer.

Assume WLOG that $d+e$ is the largest sum. So $d+e=670,$ meaning $a+b+c=2010-670=1340.$ Because we let $d+e=M,$ we must have $a+b \leq M=670$ and $b+c \leq M=670.$ Adding these inequalities gives $a+2b+c \leq 1340.$ But we just showed that $a+b+c=1340,$ which means that $b=0,$ a contradiction because we are told that all the variables are positive.

Therefore, the answer is $\boxed{B}.$

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

Contents

Problem 14

Solution 1

Solution 2

See also