2018 AIME I Problems/Problem 9
Find the number of four-element subsets of with the property that two distinct elements of a subset have a sum of , and two distinct elements of a subset have a sum of . For example, and are two such subsets.
This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set .
Note that there are only two cases: 1 where and or 2 where and . Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you , which cannot be true.
Case 1. This is probably the simplest: just make a list of possible combinations for and . We get for the first and for the second. That appears to give us solutions, right? NO. Because elements can't repeat, take out the supposed sets That's ten cases gone. So for Case 1.
Case 2. We can look for solutions by listing possible values and filling in the blanks. Start with , as that is the minimum. We find , and likewise up to . But we can't have or because or , respectively! Now, it would seem like there are values for and unique values for each , giving a total of , but that is once again not true because there are some repeated values! There are two cases of overcounting:
case 1) (5,11,13,19) & (220.127.116.11)
The same is for (6,10,14,18) and (7,9,15,17)
case 2) those that have the same b and c values
this case includes:
(1,15,9,7) and (7,9,15,1)
(2,14,10,6) and (6,10,14,2)
(3,13,11,5) and (5,11,13,3)
So we need to subtract 6 overcounts. So, that's for Case 2.
Total gives .
Let's say our four elements in our subset are . We have two cases. Note that the order of the elements / the element letters themselves don't matter since they are all on equal grounds at the start.
List out possibilities for but don't list because those are the same elements and that is restricted.
Then list out the possibilities for but don't list because they are the same elements.
This will give you elements, which is . However, as stated above, we have overlap. Just count starting from all overlap once, which is , thus cases in this case. Note that wasn't included because again, if , and cannot be .
Here, is included in both equations. We can easily see that will never equal each other.
Furthermore, there are 17 choices for ( included elements) for each . Listing out the possible s, we go from . Do not include or because if they are included, then will be the same as , which is restricted.
There are options there, and thus . But, if and , notice that . That means that if is also , then we have a double-counted set. Starting with , we have (where is . That means there are double-counted cases. Thus cases in this case.
Adding these up, we get
~IronicNinja ~ by AlcBoy1729 ~Formatted by ojaswupadhyay and phoenixfire
Solution 3 (Official MAA)
There are two types of that have the needed property. There is either an assignment of distinct values for and such that and or an assignment such that and These two types are mutually exclusive because and imply that For the first type, there are choices for namely and and there are choices for namely and Thus a four-element subset of the first type can be formed by taking the union of one of two-element subsets with one of two-element subsets as long as these two subsets are disjoint. There are such pairings that are not disjoint out of the pairings, so there are subsets of the first type.
For subsets of the second type, there are choices for the value of such that and can be two other elements of the subset. Note that in each of these cases, For each of these, there are other values that can be chosen for the element in the subset. But counts some subsets more than once. In particular, a subset is counted twice if or . In such cases either or . There are exactly subsets where the role of can be played by two different elements of the set. They are and . Thus there are subsets of the second type.
In all, there are subsets with the required property.
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