Puzzles or short exercises illustrating mathematical problem solving to freshman students

5 points are placed on a sphere. Show that there is a (closed) hemisphere which contains at least 4 points.

You can give a hint of "If we blindly take any cut of the sphere, by the pigeonhole principle, one of these hemispheres must contain at least 3 points."

Answer: Take any 2 points, and consider their great circle, which forms 2 hemispheres. By the Pigeonhole principle, out of the remaining 3 points, at least one of them must have 2 points on it. Hence we are done.

There is a thin horizontal railing, say 10 ft long, running North/South. At time $t=0$ one hundred ants fall onto the railing, at random locations and each initially facing randomly either North or South. Upon making contact with the railing all the ants begin a forward march at the speed of 10 ft per minute. The railing is thin so that whenever two ants knock antlers, they both immediately turn on their six heels and start heading in the opposite direction. Whenever an ant reaches the end of the railing, it gently floats to the ground and stays there.

How long will it take for the railing to be clear of ants in the worst case? Assume that the ants are pointlike and need zero time to turn.

No hint here. A spoiler only:

What would change, globally speaking, if the ants would be able to walk thru each other?

If no light bulb flashes quickly, one may be to try with two, three, four ants... Often a good approach to try a simpler case first.

The locker puzzle:

In a hallway there are 100 closed lockers. Now 100 persons pass the hallway. The first one changes the state (open or closed) of every locker. The second one changes the state of every second locker, the third person of every third locker and so on.

Which lockers will be open in the end?

(Answer: The lockers whose position is a perfect square.)

One can investigate this problem by doing it explicitely for numbers like 10 or 20. If some results are found, the correct answer can be conjectured. The last step is then to rigorously prove the conjecture.