2018 AIME I Problems/Problem 9

Problem

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.

Solution 1

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! We can subtract 1 from all pairs of sets that have two elements in common, because those can give us identical sets. Write out all the sets and we get . That's for Case 2.

Total gives .

-expiLnCalc

Solution 2

This is gonna be real quick, I considered making this an extension to solution one but I've decided to make it a separate solution. Disclaimer: I'm not formatting this properly, if anyone wants to format is correctly using Latex feel free.

Let's say our four elements in our subset are a,b,c,d. We have two cases. First, we can have where two distinct elements sum to 16 and two other distinct elements sum to 24. Basically, one way of representing this is having a+b = 16 and c+d = 24. 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 a+b (i.e., 1+15, 2+14, 3+13 etc.) but don't include 8+8 because those are the same elements and that is restricted. Then list out the possibilities for c+d (i.e. 4+20, 5+19, 6+18, etc.) but again don't list 12+12 because they are the same elements. This will give you 7*8 elements, which is 56. However, like above stated, we have overlap. Just count starting from a+b - 15,14,13,4,5,11,6,10,7,9 all overlap once, which is 10, thus 56 - 10 = 46 cases in this first case. Note that 12 wasn't include because again, if c+d = 24, c and d cannot be 12.

Our second case is when a+b = 16 and b+c = 24. Here, one element (b) is included twice in both equations. We can easily see that a, b, c will never equal each other. Furthermore, there are 17 choices for d (20 - 3 included elements) for each b. Listing out the possible bs, we go from 15,14,13,11,10,9,7,6,5,4. Do not include 8 or 12 because if they are included, then a/c will be the same as b, which is restricted. There are 10 options there, and since 10*17 = 170, you would think there are 170 cases and we are done.

That's the tricky part about this problem - not the computation itself, but how easily sillies can be made. This is because, if a+b = 16 and b+c = 24, notice that c-a = 8. That means that if b-d also is 8, then we have a double counted set. Starting with b=15, we have 15, 14, 13, 11, 10, 9 (where d is 7, 6, 5, 3, 2, 1). That means there are 6 double counted cases.

Adding these up, we get

~IronicNinja

Again if anyone wants to format this better feel free, I was just lazy.

Solution 0

This code works:

 int num = 0;
 for(int i = 1; i <= 20; i++){
   for(int j = i+1; j <= 20; j++){
     for(int k = j+1; k <= 20; k++){
       for(int m = k+1; m <= 20; m++){
         if(i+j==16 || i + k == 16 || i + m == 16 || j + k == 16
         || j + m == 16 || k + m == 16){
           if(i+j==24 || i+k==24 || i+m==24 || j+k==24 || j+m == 24 || k+m==24){
             num++; 
           }
         }
       }
     }
   }
 }
 cout << num << endl;

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.