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

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

Revision as of 22:14, 18 January 2020

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 $(w,x,y,z)=(16, 11, 10, 7)$ 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 $15 + 16 =$16+15=\boxed{\textbf{(B) }31}$$ (Error compiling LaTeX. Unknown error_msg).

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 $\boxed{31}$. You can check that 15 and 16 work by forming a string of 4 numbers as shown above.

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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