2023 AMC 8

Solution 1

First, we need to find the sum of each group when split. This is the total sum of all the elements divided by the # of groups. . Then, dividing by , we have , so each group of must have a sum of 15. To make the counting easier, we will just see the possible groups 9 can be with. The possible groups 9 can be with 2 distinct numbers are and . Going down both of these avenues, we will repeat the same process for using the remaining elements in the list. Where there is only 1 set of elements getting the sum of , needs in both cases. After is decided, the remaining 3 elements are forced in a group, yielding us an answer of as our sets are and .

Solution 2

The group with must have the two other numbers adding up to , since the sum of all the numbers is . The sum of the numbers in each group must therefore be . We can have , , , or . With the first group, we have left over. The only way to form a group of numbers that add up to is with or . One of the possible arrangements is therefore . Then, with the second group, we have left over. With these numbers, there is no way to form a group of numbers adding to . Similarly, with the third group there is left over and we can make a group of numbers adding to with or . Another arrangement is . Finally, the last group has left over. There is no way to make a group of numbers adding to with this, so the arrangements are and . So, there are sets that can be formed.