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Difference between revisions of "2011 AMC 10B Problems/Problem 21"
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w&=15 \end{align*}</cmath>
 
w&=15 \end{align*}</cmath>
 
The sum of the possible values of <math>w</math> is <math>16+15 = \boxed{\textbf{(B) }31}</math>
 
The sum of the possible values of <math>w</math> is <math>16+15 = \boxed{\textbf{(B) }31}</math>
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 +
== Solution 2 ==
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First, like Solution 1, we know that <math>w-z=9 \ \text{(1)}</math>, because no sum could be smaller. Next, we find the sum of all the differences; since <math>w</math> is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes <math>3w</math>. Continuing in this way, we find that <cmath>3w+x-y-3z=28 \ \text{(2)}</cmath>. Now, we can subtract <math>3w-3z=27</math> from (2) to get <math>x-y=1 \ \text{(3)}</math>. Also, adding (2) with <math>w+x+y+z=44</math> gives <math>4w+2x-2z=72</math>, or <math>2w+x-z=36</math>. Subtracting (1) from this gives <math>w+z=27</math>. Since we know <math>w-z</math> and <math>x-y</math>, we find that <cmath>(w-z)+(x-y)=(w-y)+(x-z)=9+1=10</cmath>. This means that <math>w-y</math> and <math>x-z</math> must be 4 and 6, in some order. If <math>w-y=6</math>, then subtracting this from (3) gives <math>(w-y)-(x-y)=6-1=5</math>, so <math>w-x=5</math>. This means that <math>(w-x)+(w+x)=2w=27+5=32</math>, so <math>w=16</math>. Similarly, <math>w</math> can also equal <math>15</math>.
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Now if you are in a rush, you would have just answered <math>16+15=\boxed{\boxed{\textbf{(B) }31}}</math>. But we do have to check if these work. In fact, they do, giving solutions <math>\boxed{16, 11, 10, 7}</math> and <math>\boxed{15, 12, 11, 6}</math>.
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<math>\blacksquare</math>
  
 
== See Also==
 
== See Also==

Revision as of 11:03, 15 November 2013

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 for $w$?

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

Solution

The largest difference, $9,$ must be between $w$ and $z.$

The smallest difference, $1,$ must be directly between two integers. This also means the differences directly between the other two should add up to $8.$ The only remaining differences that would make this possible are $3$ and $5.$ However, those two differences can't be right next to each other because they would make a difference of $8.$ This means $1$ must be the difference between $y$ and $x.$ We can express the possible configurations as the lines.


[asy] unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1); pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1); draw(Z1--W1); draw(Z4--W4); pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4}; dot(ps); label("$z$",Z1,N); label("$y$",Y1,N); label("$x$",X1,N); label("$w$",W1,N); label("$z$",Z4,N); label("$y$",Y4,N); label("$x$",X4,N); label("$w$",W4,N); label("$1$",(X1--Y1),N); label("$1$",(X4--Y4),N); label("$3$",(Y1--Z1),N); label("$3$",(W4--X4),N); label("$5$",(X1--W1),N); label("$5$",(Y4--Z4),N); [/asy]

If we look at the first number line, you can express $x$ as $w-5,$ $y$ as $w-6,$ and $z$ as $w-9.$ Since the sum of all these integers equal $44$, \begin{align*} w+w-5+w-6+w-9&=44\\ 4w&=64\\ w&=16 \end{align*} You can do something similar to this with the second number line to find the other possible value of $w.$ \begin{align*} w+w-3+w-4+w-9&=44\\ 4w&=60\\ w&=15 \end{align*} The sum of the possible values of $w$ is $16+15 = \boxed{\textbf{(B) }31}$

Solution 2

First, like Solution 1, we know that $w-z=9 \ \text{(1)}$, because no sum could be smaller. Next, we find the sum of all the differences; since $w$ is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes $3w$. Continuing in this way, we find that \[3w+x-y-3z=28 \ \text{(2)}\]. Now, we can subtract $3w-3z=27$ from (2) to get $x-y=1 \ \text{(3)}$. Also, adding (2) with $w+x+y+z=44$ gives $4w+2x-2z=72$, or $2w+x-z=36$. Subtracting (1) from this gives $w+z=27$. Since we know $w-z$ and $x-y$, we find that \[(w-z)+(x-y)=(w-y)+(x-z)=9+1=10\]. This means that $w-y$ and $x-z$ must be 4 and 6, in some order. If $w-y=6$, then subtracting this from (3) gives $(w-y)-(x-y)=6-1=5$, so $w-x=5$. This means that $(w-x)+(w+x)=2w=27+5=32$, so $w=16$. Similarly, $w$ can also equal $15$.

Now if you are in a rush, you would have just answered $16+15=\boxed{\boxed{\textbf{(B) }31}}$. But we do have to check if these work. In fact, they do, giving solutions $\boxed{16, 11, 10, 7}$ and $\boxed{15, 12, 11, 6}$.

$\blacksquare$

See Also

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 10 Problems and Solutions

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