Problem:
Alina writes the numbers on separate cards, one number per card. She wishes to divide the cards into groups of cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
The sum of all integers is , so the sum of the integers in each group must be . Because is odd, the number of odd integers in each group must be odd, so one group must contain three odd integers and the other two must contain one each. The only groups of three odd integers with a sum of are {} and {}.
If the group of three odd integers is {}, then the number must be combined with two even integers with a sum of , and the only possibility is {} leaving {} as the third group.
If the group of three odd integers is {}, then the number must be combined with two even integers with a sum of , and the only possibility is {}, leaving {} as the third group.
Therefore the cards can be divided into groups with identical sums in exactly ways.
The sum of all nine integers is , so the sum of the integers in each group must be . Observe that the numbers , and must all be in different groups, because if two of them were in the same group, the sum of that group would be too large. Symmetrically, the numbers , and must all be in different groups, because if two of them were in the same group, with any third number the sum of that group would be too small. Consider the group that contains the number . The two remaining numbers add up to , so the group must be {}, {}, or {}. (Note that the number cannot be grouped with and because each group must contain , or .) If {} is one of the groups, then the other two groups must be {} and {}. Similarly, if {} is one of the groups, then the other two groups must be {} and {}. There is no way for {} to be one of the groups, because both of the ways to match and with and to make groups of sum result in repeated numbers. Therefore the cards can be divided into groups with identical sums in exactly ways.
Answer: .
The problems on this page are the property of the MAA's American Mathematics Competitions