2011 AMC 10B Problems/Problem 21

Problem

Brian writes down four integers Lizzie writes down the opposites of the integers Brian wrote down. If the product of the integers Brian wrote down is 756 what is the product of the integers Lizzie wrote down

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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