Difference between revisions of "2011 AMC 10B Problems/Problem 21"
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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> | ||
− | == Solution 2 == | + | === 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>. |
Revision as of 21:40, 18 January 2016
Contents
Problem
Brian writes down four integers whose sum is . The pairwise positive differences of these numbers are and . What is the sum of the possible values for ?
Solution
The largest difference, must be between and
The smallest difference, must be directly between two integers. This also means the differences directly between the other two should add up to The only remaining differences that would make this possible are and However, those two differences can't be right next to each other because they would make a difference of This means must be the difference between and We can express the possible configurations as the lines.
If we look at the first number line, you can express as as and as Since the sum of all these integers equal , You can do something similar to this with the second number line to find the other possible value of The sum of the possible values of is
Solution 2
First, like Solution 1, we know that , because no sum could be smaller. Next, we find the sum of all the differences; since 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 . Continuing in this way, we find that . Now, we can subtract from (2) to get . Also, adding (2) with gives , or . Subtracting (1) from this gives . Since we know and , we find that . This means that and must be 4 and 6, in some order. If , then subtracting this from (3) gives , so . This means that , so . Similarly, can also equal .
Now if you are in a rush, you would have just answered . But we do have to check if these work. In fact, they do, giving solutions and .
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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