1983 AIME Problems/Problem 13
Contents
Problem
For and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for is and for it is simply . Find the sum of all such alternating sums for .
Solution
Solution 1
Let be a non- empty subset of .
Then the alternating sum of , plus the alternating sum of , is . This is because, since is the largest element, when we take an alternating sum, each number in ends up with the opposite sign of each corresponding element of .
Because there are of these pairs of sets, the sum of all possible subsets of our given set is , giving an answer of .
Solution 2 (almost the same as Solution 1)
Consider a given subset of that contains ; then there is a subset which contains all the elements of except for , and only those elements . Since each element of has one fewer element preceding it than it does in , their signs are opposite. Thus the sum of the alternating sums of and is equal to 7. There are subsets containing 7, so our answer is .
Solution 3
Denote the desired total of all alternating sums of an -element set as . We are looking for . Notice that all alternating sums of an -element set are also alternating sums of an -element set. However, when we go from an to element set, for each subset with the new element, we are adding the new element and subtracting one of the alternating sums of the -element set. There are subsets of an -element set that includes the new element, giving us the relationship . When , we therefore get .
Solution 4
We analyze all the numbers from 1 to 7 separately to see where the number contributes its positive or negative to the sum of the alternating sums. Whenever 7 appears, which it does 64 times, it contributes a positive because it is always first. This gives a net gain of .
If we look at when 6 appears, which it also does 64 times, whether it comes as positive or negative depends on the presence of 7. Half of the subsets with 6 have 7 resulting in subtracting 6 each time, while the other half does not have 7 adding 6 each time, so these contributions of sixes cancel each other out giving a net gain of 0.
The same thing happens to any positive integer less than 7. This is because the determination of a positive or negative contribution is dependent on the number of larger numbers in front of it(For example, the sign of 3 is dependent on the presence of 4, 5, 6, and 7 in the subset). If the number of larger numbers is even, it gives in a positive copy while odd produces its negative. We know that the frequencies of these two cases occurring are the same because via the Binomial Theorem. Therefore, all positive integers less than 7 will not have any effect and our sum will be .
Solution 5
Let . Let the alternating sum of a certain subset of of be and let We see that as if is the largest element in Now, we know that so Thus, our answer (which is the case) is
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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