[Math] Puzzles or short exercises illustrating mathematical problem solving to freshman students

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At high school, the solution method to almost all mathematical exercises is to apply some technique or algorithm you have learned before. At the university, the situation is fundamentally different. Many exercises have the flavor of little research problems, where it is not obvious at all how to start. This new experience can be frustrating even to the most talented students.

I'm looking for puzzles or small exercises to illustrate "mathematical problem solving" to freshman students. Ideally, such a problem statement meets the following requirements:

  • Easy to understand. The involved notions should be familiar from high school mathematics or daily live, and not be built upon a framework of abstract definitions. Suitable problems often come from combinatorics, elementary geometry or elementary number theory. Some abstractness is ok, since this is another point freshman students have to get used to.

  • Exciting or fascinating in some way. No dry theory questions.

  • The question should be suitable to get your hands on and "do reasearch" with it (if necessary, by giving hints). For example by looking at special cases, investigating the consequences of modified requirements, etc.

  • The solution should not be obvious and require some clever idea. Furthermore, it should be possible to give a rigorous solution/proof, without involving long calculations, and like the question, only relying on methods which are well-known to a freshman student.

I will some give some examples of good problems (in my opinion) as answers below.

Please give only one problem statement per answer.

Best Answer

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.

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