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Difference between revisions of "2011 AMC 12B Problems/Problem 13"

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The sum of the two w's is <math>15+16=31</math> <math>\boxed{B}</math>
 
The sum of the two w's is <math>15+16=31</math> <math>\boxed{B}</math>
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== Solution 2 ==
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Let the four numbers be <math>z</math>, <math>z+a</math>, <math>z+b</math>, and <math>z+c</math>. We know that <math>c</math> must be <math>9</math> because that's the greatest difference. So we have <math>z</math>, <math>z+a</math>, <math>z+b</math>, and <math>z+9</math>. The 6 possible differences are <math>a</math>, <math>b</math>, <math>9</math>, <math>b-a</math>, <math>9-a</math>, and <math>9-b</math>. We are given that the differences are 1, 3, 4, 5, 6, 9. <math>a</math> and <math>9-a</math> and <math>b</math> and <math>9-b</math> add to 9 which means they have to be 4, 5 and 3,6 or vice versa. Which leaves <math>1</math>. That means <math>b-a</math> has to equal <math>1</math>. So an and b have to be 3,4, 4,5, or 5,6. For 3,4, we have <math>z</math>, <math>z+3</math>, <math>z+4</math>, and <math>z+9</math>. <math>4z+16=44</math>. <math>4z=28</math>. <math>z=7</math>. <math>w=7+9=16</math>. Now for 4,5, notice that it doesn't work. The differences are 4, 5, 9, 1, 4, 5. We are missing 6 and 3. For 5,6, it's <math>z</math>, <math>z+5</math>, <math>z+6</math>, and <math>z+9</math>. Check that the differences work; they do. We have <math>4z+20=44</math>. <math>4z=24</math>. <math>z=6</math>. <math>w=6+9=15</math>. Therefore our answer is <math>15+16=\boxed{31}</math>.
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~MC413551
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|ab=B|num-b=12|num-a=14}}
 
{{AMC12 box|year=2011|ab=B|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:55, 8 August 2023

Problem

Brian writes down four integers $w > x > y > z$ whose sum is $44$. The pairwise positive differences of these numbers are $1, 3, 4, 5, 6$ and $9$. What is the sum of the possible values of $w$?

$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 93$

Solution

Assume that $y-z=a, x-y=b, w-x=c.$ $w-z$ results in the greatest pairwise difference, and thus it is $9$. This means $a+b+c=9$. $a,b,c$ must be in the set ${1,3,4,5,6}$. The only way for 3 numbers in the set to add up to 9 is if they are $1,3,5$. $a+b$, and $b+c$ then must be the remaining two numbers which are $4$ and $6$. The ordering of $(a,b,c)$ must be either $(3,1,5)$ or $(5,1,3)$.


\begin{align*} z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\ 4z + a + (a + b) + 9 &= 44\\ \text{if} \hspace{1cm} a &= 3 \\ a + b &= 4\\ 4z &= 44 - 9 - 3 - 4\\ z &= 7\\ w &= 16\\ \end{align*}

\begin{align*} \text{if} \hspace{1cm} a &= 5\\ a + b &= 6\\ 4z &= 44 - 9 - 5 - 6\\ z &= 6\\ w &= 15\\ \end{align*}


The sum of the two w's is $15+16=31$ $\boxed{B}$

Solution 2

Let the four numbers be $z$, $z+a$, $z+b$, and $z+c$. We know that $c$ must be $9$ because that's the greatest difference. So we have $z$, $z+a$, $z+b$, and $z+9$. The 6 possible differences are $a$, $b$, $9$, $b-a$, $9-a$, and $9-b$. We are given that the differences are 1, 3, 4, 5, 6, 9. $a$ and $9-a$ and $b$ and $9-b$ add to 9 which means they have to be 4, 5 and 3,6 or vice versa. Which leaves $1$. That means $b-a$ has to equal $1$. So an and b have to be 3,4, 4,5, or 5,6. For 3,4, we have $z$, $z+3$, $z+4$, and $z+9$. $4z+16=44$. $4z=28$. $z=7$. $w=7+9=16$. Now for 4,5, notice that it doesn't work. The differences are 4, 5, 9, 1, 4, 5. We are missing 6 and 3. For 5,6, it's $z$, $z+5$, $z+6$, and $z+9$. Check that the differences work; they do. We have $4z+20=44$. $4z=24$. $z=6$. $w=6+9=15$. Therefore our answer is $15+16=\boxed{31}$. ~MC413551

See also

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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