There's a very difficult puzzle in the latest scientific american, it goes as follows:
Three of the nine members of the president's cabinet are leaks. If the president gives a tip to some subset of his advisors and all three interlopers are in that subset, then that tip will get leaked to the press. The president has decided to determine who the leaks are by selectively giving tips to advisors and seeing which ones leak. How can this be done by giving each tip to three or four people, having no more than two leaked tips, and using no more than 25 tips total?
There's a solution on the web site, although the one I figured out is very different.
Here's my solution. First, note that if a tip to four people leaks, the leaks can be found using only three more tips, giving each of them to three of the four.
Arrange eight of the nine advisors on the vertices of a 2x2x2 cube. Test each of the 12 subsets which are coplanar, plus the 2 subsets which are all the same color if they're colored like a checkerboard. If any of those leak, we're done. If not, we've determined that the odd one out must be one of the leaks.
Next, arrange the eight we previously had in a cube formation into a line, and test all 8 configurations of the form the odd one out, x, x+1, and x+3 (modulo 8).
If none of those hit, then the two other leaks must be of the form x and x + 4, which there are four possibilities for. Three of them can be tried directly and if none leak then the last one can be inferred.
This leads to a total of 14 + 8 + 3 = 25 tips. It's a very hard problem, it took me about 45 minutes to figure out a solution.
An additional question given is whether increasing the number of advisors included in a tip can reduce the number of trials necessary. The answer is yes. For example, to modify the technique I gave the first test could be of six of the corners of the cube except for two adjacent ones. This effectively does three tests at once, so the total number of tests needed drops to 23. If the first test turns up positive, all but one of the 20 subsets of 3 of 6 can be tried individually, and if none of them leak then the last one can be inferred, for a total of one initial tip with to 6 plus 19 others, or 20 total, so leaking on the first tip isn't the limiting case.