Difference between revisions of "1983 AIME Problems/Problem 13"
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Denote the desired total of all alternating sums of an <math>n</math>-element set as <math>S_n</math>. We are looking for <math>S_7</math>. Notice that all alternating sums of an <math>n</math>-element set are also alternating sums of an <math>n+1</math>-element set. However, when we go from an <math>n</math> to <math>n+1</math> element set, for each subset with the new element, we are adding the new element and subtracting one of the alternating sums of the <math>n</math>-element set. There are <math>2^n</math> subsets of an <math>n+1</math>-element set that includes the new element, giving us the relationship <math>S_{n+1} = S_n + 2^n(n+1) - S_n = 2^n(n+1)</math>. When <math>n = 6</math>, we therefore get <math>S_ 7 = 2^6(7) = \boxed{448}</math>. | Denote the desired total of all alternating sums of an <math>n</math>-element set as <math>S_n</math>. We are looking for <math>S_7</math>. Notice that all alternating sums of an <math>n</math>-element set are also alternating sums of an <math>n+1</math>-element set. However, when we go from an <math>n</math> to <math>n+1</math> element set, for each subset with the new element, we are adding the new element and subtracting one of the alternating sums of the <math>n</math>-element set. There are <math>2^n</math> subsets of an <math>n+1</math>-element set that includes the new element, giving us the relationship <math>S_{n+1} = S_n + 2^n(n+1) - S_n = 2^n(n+1)</math>. When <math>n = 6</math>, we therefore get <math>S_ 7 = 2^6(7) = \boxed{448}</math>. | ||
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+ | === 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 <math>7 \cdot 64=448</math>. | ||
+ | |||
+ | 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. | ||
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+ | 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 <math>0=(1-1)^{n}=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-...\binom{n}{n}</math> via the Binomial Theorem. Therefore, all positive integers less than 7 will not have any effect and our sum will be <math>\boxed{448}</math>. | ||
== See Also == | == See Also == |
Revision as of 00:32, 27 December 2019
Contents
Problem
For and each of its non-empty subsets a unique alternating sum 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 (subtracting
to exclude the empty set), the sum of all possible subsets of our given set is
. However, we forgot to include the subset that only contains
, so the answer is
.
Note: The empty set is the same as the subset that only includes so you could have just left it as
pairs of sets.
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
.
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |