2011 AMC 10B Problems/Problem 21

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 $w$ ?

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 $8,$ which isn't given as a possibility in the problem. 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 two numbers could have a larger difference. 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 most likely would have answered $16+15=\boxed{\textbf{(B) }31}$ . But we do have to check if these work. In fact, they do, giving solutions $(w,x,y,z)=(16, 11, 10, 7)$ and $(w,x,y,z)=(15, 12, 11, 6)$ .

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 $(a, b, c) = (1, 6, 9), (3, 4, 9),$ 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 $16+15=\boxed{\textbf{(B) }31}$ .

Solution 4

From the problem, we know that $w+x+y+z=44$ . Since it is said that the pairwise positive differences between numbers are 1, 3, 4, 5, 6, 9, and we can figure that the pairwise positive differences are $(w-x)$ , $(w-y)$ , $(w-z)$ , $(x-y)$ , $(x-z)$ , $(y-z)$ , the sum of $(w-x), (w-y), (w-z), (x-y), (x-z), (y-z)$ is equal to the sum of 1, 3, 4, 5, 6, 9, so $(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28$ . Simplifying, we get $3w-3z+x-y=28$ . Adding $w+x+y+z=44$ and $(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28$ , we get $4w+2x-2z=72$ , and simplifying we get $2w+x-z=36$ . Since $x-z$ is one of our positive differences, we can start guessing values for $w$ , and if the equation simplifies to one of our numerical positive differences, that value of $w$ should work. We can start at $w=18$ and keep going down, because our sum has to be positive. For $w=18$ , $x-z=0$ , which is not one of our sums. For $w=17$ , $x-z=2$ ,which is not one of our sums. For $w=16$ , $x-z=4$ , which is one of our sums, so 16 works. For $w=15$ , $x-z=6$ , which is one of our sums, so 15 works. For $w=14$ , $x-z=8$ , which is not one of our sums. If we keep going, $x-z$ will soon exceed 10 and exceed all our sums, so any value below $w=14$ will not work. Therefore, our only solutions for $w$ are 15 and 16, which means our sum is $\boxed{\textbf{(B) }31}$ . You can check that 15 and 16 work by forming a string of 4 numbers as shown above.

Solution 5

Because we know that $w>x>y>z$ and that the positive differences are $1, 3, 4, 5, 6, 9$ , we can immediately come to the conclusion that $w-z= 9$ (because w is the largest integer and z is the smallest integer, so their difference must be the greatest). With this we have $3$ equations, we have that $x+y+z+w=44$ (from the problem), $w-z=9$ and because we can add up all the possible possible differences (as shown in the previous solutions), we get that $3w+x-y-3z=28$ . With these equations, we eventually manipulate these equations by doing the first equation minus 3 times the second to get $x-y=1$ . We can also add the first and third equation and subtract the second equation to get $w+x=27$ thus we know that $w-x$ can be $3, 4, 5,$ and $6$ (note: 1 and 9 are not possible because we know that $x-y=1$ and $w-z=9$ and the remaining differences can only be taken by 1 pair, so $w-x$ cannot be equal to 1 or 9). However, in order to get an integer value for x and z, we find that $w-z$ can only be equal to 3 and 5. Thus, by solving these, we see that $w= 15$ and $w=16$ . $15+16=\boxed{\textbf{(B) }31}$ .

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
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