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

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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 \Rightarrow \boxed{B}</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>.
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== Solution 2 ==
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First, note that, simply by pigeonhole, at least one of a, b, c, d, e is greater than or equal to <math>\frac{2010}{5}=402,</math> 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.
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Assume WLOG that <math>d+e</math> is the largest sum. So <math>d+e=670,</math> meaning <math>a+b+c=2010-670=1340.</math> Because we let <math>d+e=M,</math> we must have <math>a+b \leq M=670</math> and <math>b+c \leq M=670.</math> Adding these inequalities gives <math>a+2b+c \leq 1340.</math> But we just showed that <math>a+b+c=1340,</math> which means that <math>b=0,</math> a contradiction because we are told that all the variables are positive.
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Therefore, the answer is <math>\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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:17, 1 December 2016

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

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 $\boxed{B}.$

See also

2010 AMC 12B (ProblemsAnswer KeyResources)
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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