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>.
  
== Solution 2 ==
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== Solution 2 ==
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.
 
 
 
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.
 
 
 
Therefore, the answer is <math>\boxed{B}.</math>
 
 
 
=== Solution 3 ===
 
 
Since <math>a + b \le M</math>, <math>d + e \le M</math>, and <math>c < b + c \le M</math>, we have that <math>2010 = a + b + c + d + e < 3M</math>. Hence, <math>M > 670</math>, or <math>M \ge 671</math>.
 
Since <math>a + b \le M</math>, <math>d + e \le M</math>, and <math>c < b + c \le M</math>, we have that <math>2010 = a + b + c + d + e < 3M</math>. Hence, <math>M > 670</math>, or <math>M \ge 671</math>.
  
 
For the values <math>(a,b,c,d,e) = (669,1,670,1,669)</math>, <math>M = 671</math>, so the smallest possible value of <math>M</math> is <math>\boxed{671}</math>. The answer is (B).
 
For the values <math>(a,b,c,d,e) = (669,1,670,1,669)</math>, <math>M = 671</math>, so the smallest possible value of <math>M</math> is <math>\boxed{671}</math>. The answer is (B).
 +
 +
~ math31415926535
  
 
== 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}}

Latest revision as of 23:30, 14 January 2021

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

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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