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Difference between revisions of "2011 AMC 10B Problems/Problem 21"

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<math> \textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93</math>
 
<math> \textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93</math>
  
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== Solution ==
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== Solution 1 ==
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=== Solution 1 ===
 
 
The largest difference, <math>9,</math> must be between <math>w</math> and <math>z.</math>
 
The largest difference, <math>9,</math> must be between <math>w</math> and <math>z.</math>
  
 
The smallest difference, <math>1,</math> must be directly between two integers. This also means the differences directly between the other two should add up to <math>8.</math> The only remaining differences that would make this possible are <math>3</math> and <math>5.</math> However, those two differences can't be right next to each other because they would make a difference of <math>8.</math> This means <math>1</math> must be the difference between <math>y</math> and <math>x.</math> We can express the possible configurations as the lines.
 
The smallest difference, <math>1,</math> must be directly between two integers. This also means the differences directly between the other two should add up to <math>8.</math> The only remaining differences that would make this possible are <math>3</math> and <math>5.</math> However, those two differences can't be right next to each other because they would make a difference of <math>8.</math> This means <math>1</math> must be the difference between <math>y</math> and <math>x.</math> We can express the possible configurations as the lines.
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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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== Solution 2 ==
  
 
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+x=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>.
 
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+x=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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=== Solution 3 === (ccx09)
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== Solution 3 ==
  
 
Let w - x = a, w - y = b, w - z = c. As above, we know that c = 9. Thus, a < b < c.  
 
Let w - x = a, w - y = b, w - z = c. As above, we know that c = 9. Thus, a < b < c.  

Revision as of 20:54, 4 February 2017

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 1

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 $w+x=27$. 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 $16+15=\boxed{\textbf{(B) }31}$. But we do have to check if these work. In fact, they do, giving solutions $\boxed{16, 11, 10, 7}$ and $\boxed{15, 12, 11, 6}$.


Solution 3

Let w - x = a, w - y = b, w - z = c. As above, we know that c = 9. Thus, a < b < c. So, we have w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44. This means a + b + 9 is a multiple of 4. Testing values of a and b, we find (a, b, c) = (1, 6, 9), (3, 4, 5), and (5, 6, 9) all satisfy this relation. The corresponding (w, x, y, z) sets are (15, 14, 9, 6), (15, 12, 11, 6), and (16, 11, 10, 7). The first set does not satisfy the given conditions, but the other two do. Thus, w = 15 and w = 16 are both possible solutions so the answer is 15 + 16 = 31.

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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