Problem:
Harold, Tanya, and Ulysses paint a very long picket fence.
Harold starts with the first picket and paints every picket;
Tanya starts with the second picket and paints every picket; and Ulysses starts with the third picket and paints every picket.
Call the positive integer paintable when the triple of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers.
Solution:
Let the pickets be numbered consecutively Let , and be the sets of numbers assigned to the pickets painted by Harold, Tanya, and Ulysses, respectively. Then
Each picket will be painted exactly once if and only if , and partition the set of positive integers into mutually disjoint subsets. Clearly, and are each greater than . In fact, , since if , then is in . Also , because if , then cannot be in ; since cannot be in would have to be in , making , which would make , since is not in . But this leaves no possible value for . Thus or . When , . Now cannot be in because would be too, but is in . So is in , and , which means that . When , . Since cannot be in , so . The two paintable integers are and , whose sum is .
The problems on this page are the property of the MAA's American Mathematics Competitions