2011 AMC 10B Problems/Problem 21
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 1
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 which isn't given as a possibility in the problem. 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 two numbers could have a larger difference. 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 most likely would have answered . But we do have to check if these work. In fact, they do, giving solutions and .
Solution 3
Let , , . As above, we know that . Thus, . So, we have . This means is a multiple of . Testing values of and , we find and all satisfy this relation. The corresponding sets are and . The first set does not satisfy the given conditions, but the other two do. Thus, and are both possible solutions so the answer is .
Solution 4
From the problem, we know that . 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 , , , , , , the sum of is equal to the sum of 1, 3, 4, 5, 6, 9, so . Simplifying, we get . Adding and , we get , and simplifying we get . Since is one of our positive differences, we can start guessing values for , and if the equation simplifies to one of our numerical positive differences, that value of should work. We can start at and keep going down, because our sum has to be positive. For , , which is not one of our sums. For , ,which is not one of our sums. For , , which is one of our sums, so 16 works. For , , which is one of our sums, so 15 works. For , , which is not one of our sums. If we keep going, will soon exceed 10 and exceed all our sums, so any value below will not work. Therefore, our only solutions for are 15 and 16, which means our sum is . You can check that 15 and 16 work by forming a string of 4 numbers as shown above.
Solution 5
Because we know that and that the positive differences are , we can immediately come to the conclusion that (because w is the largest integer and z is the smallest integer, so their difference must be the greatest). With this we have equations, we have that (from the problem), and because we can add up all the possible possible differences (as shown in the previous solutions), we get that . With these equations, we eventually manipulate these equations by doing the first equation minus 3 times the second to get . We can also add the first and third equation and subtract the second equation to get thus we know that can be and (note: 1 and 9 are not possible because we know that and and the remaining differences can only be taken by 1 pair, so cannot be equal to 1 or 9). However, in order to get an integer value for x and z, we find that can only be equal to 3 and 5. Thus, by solving these, we see that and . .
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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