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Difference between revisions of "2010 AMC 12B Problems/Problem 14"

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We want to try make <math>a+b</math>, <math>b+c</math>, <math>c+d</math>, and <math>d+e</math> as close as possible so that <math>M</math>, the maximum of these, if smallest.
 
We want to try make <math>a+b</math>, <math>b+c</math>, <math>c+d</math>, and <math>d+e</math> as close as possible so that <math>M</math>, the maximum of these, if smallest.
  
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Notice that <math>2010=670+670+670</math>. In order to express <math>2010</math> as a sum of <math>5</math> 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): <math>2010=670+1+670+1+668</math> or <math>2010=670+1+669+1+669</math>. We see that in both cases, the value of <math>M</math> is <math>671</math>, so the answer is <math>671</math>.
+
Notice that <math>2010=670+670+670</math>. In order to express <math>2010</math> as a sum of <math>5</math> 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): <math>2010=670+1+670+1+668</math> or <math>2010=670+1+669+1+669</math>. We see that in both cases, the value of <math>M</math> is <math>671</math>, so the answer is <math>671 \Rightarrow \boxed{B}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=13|num-a=15|ab=B}}
 
{{AMC12 box|year=2010|num-b=13|num-a=15|ab=B}}

Revision as of 12:55, 31 May 2011

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 $671 \Rightarrow \boxed{B}$.

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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