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

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== Problem 14 ==
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== Problem ==
 
Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> be positive integers with <math>a+b+c+d+e=2010</math> and let <math>M</math> be the largest of the sum <math>a+b</math>, <math>b+c</math>, <math>c+d</math> and <math>d+e</math>. What is the smallest possible value of <math>M</math>?
 
Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> be positive integers with <math>a+b+c+d+e=2010</math> and let <math>M</math> be the largest of the sum <math>a+b</math>, <math>b+c</math>, <math>c+d</math> and <math>d+e</math>. What is the smallest possible value of <math>M</math>?
  
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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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== 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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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>.
  
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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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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).
  
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Therefore, the answer is <math>\boxed{B}.</math>
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~ math31415926535
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==Solution 3==
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Notice that only the sums of adjacent numbers matter. (For example, a & c could be extremely high, as long as b is relatively low.)
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Therefore creating "mountains" and "valleys" is the best way to lower the sum of adjacent numbers. We can do
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1. (high, low, high, low, high)
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or
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2. (low, high, low, high, low)
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 +
 
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In the extreme case that each "low" = <math>0</math>, <math>2010</math> will be divided into either <math>3</math> or <math>2</math> numbers for cases 1 and 2, respectively. Obviously dividing by <math>3</math> will yield a lower number, so we consider case 1.
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Dividing <math>2010</math> by <math>3</math> yields <math>670</math>, or
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(<math>670, 0, 670, 0, 670</math>)
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However, all five numbers must be positive, and the closest we can get to this is
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(<math>668, 3, 668, 3, 668</math>)
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The lowest possible sum of two adjacent numbers then becomes <math>671</math>, or <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}}
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[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:38, 17 April 2024

Problem

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

Solution 3

Notice that only the sums of adjacent numbers matter. (For example, a & c could be extremely high, as long as b is relatively low.) Therefore creating "mountains" and "valleys" is the best way to lower the sum of adjacent numbers. We can do


1. (high, low, high, low, high)


or


2. (low, high, low, high, low)


In the extreme case that each "low" = $0$, $2010$ will be divided into either $3$ or $2$ numbers for cases 1 and 2, respectively. Obviously dividing by $3$ will yield a lower number, so we consider case 1.

Dividing $2010$ by $3$ yields $670$, or


($670, 0, 670, 0, 670$)


However, all five numbers must be positive, and the closest we can get to this is


($668, 3, 668, 3, 668$)


The lowest possible sum of two adjacent numbers then becomes $671$, or $\boxed{B}$.

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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