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

Solution 2

Since $a + b \le M$ , $d + e \le M$ , and $c < b + c \le M$ , we have that $2010 = a + b + c + d + e < 3M$ . Hence, $M > 670$ , or $M \ge 671$ .

For the values $(a,b,c,d,e) = (669,1,670,1,669)$ , $M = 671$ , so the smallest possible value of $M$ is $\boxed{671}$ . The answer is (B).

~ math31415926535

2010 AMC 12B (Problems • Answer Key • Resources)
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2010 AMC 12B Problems/Problem 14

Contents

Problem 14

Solution 1

Solution 2

See also