So, this turns out to be very straightforward.  The problem is that it appears to be intimidating to lots of students because they don’t have a firm grasp algebraically on the concept of a prime number.  And this, to me at least, seems like the major stumbling block on all of number theory, which is why it is such a fertile source of good contest problems.  You will never have a firm algebraic grasp of a prime number (ignoring of course the rather large subject of algebraic number theory), so it suddenly seems that all of the high school training in algebra is no good here.  Nevertheless, the main way to solve problems is to actually attempt to solve them instead of throwing up your hands and bathering in despair like a souless zombie from Left 4 Dead Two.

So, how to actually tackle this problem?

Consider that a positive prime number has as its divisors just and .  This means that the number can only have divisors , , , , , , and .  This means that the sum of the divisors is

.

In hindsight, not too bad at all.  When I’ve shown this to my students after letting them struggle for awhile, they usually say “Oh, well that’s pretty easy” and unfortunately don’t seem to learn from the problem.  I like to review the thinking process that students used to solve the problem, and see what they learn from their attempts.  Usually, the ones who are successful tried several attempts, including putting in small values for to see if that helped with a solution.  I have also found that the successful ones stuck with it for longer than a minute, unless of course, a solution was found in less time.

This problem came from Problems and How to Solve Them Volume 1, which is a nice little manual put out by the Centre for Education in Mathematics and Computing at the University of Waterloo on how to help high school students with learning problems solving.  It is set as problem 15 of Chapter 6, Properties of Integers, Divisibility.  Over time, I will probably share a few more problems from the CEMC, mainly because they run a number of excellent contests for Canada that eventually lead to IMO-level of difficulty, but also start in the elementary grades, so there is a full range of challenges for students and teachers alike.

To combat this of course, we need to expose the students to as many problems as possible, giving them time to solve them, but not so long that they abandon the subject.  Here are a few that I find interesting.  One easy, the other more challenging.

One: If is a prime number greater than , find the sum of the divisors of .

Two: If and are positive integers such that , show that is divisible by .

The key to this problem is that we are just writing down the numbers themselves, which means that we need to write the numbers in such a way as to create a large number.  At first, it seems that the largest number is , which of course can’t be right, since that is too obvious (and also incorrect).  In order to make any headway, it is necessary to realize that exponents don’t require anything except placement of numbers in a different part of the arithmetic expression.  So the next obvious ploy is to use exponents.  By checking different permutations of the numbers in the base and exponent, we see that is the largest of the set under consideration.  It has an impressive digits.  (Incidentally, was a close second here, with an equally impressive digits.

As it turns out, this is also not quite enough.  The real key to this problem is to use exponents of exponents.  At this point, we need to get the exponent very large, which (as long as the base is not one) will create a genuinely massive number.  The one to do it is , which has a total of about digits.  Quite  the number.

I first saw this particular problem in print in Pickover’s Wonders of Numbers, although I had come across discussions of something similar from my undergraduate days when learning about tower powers.  I definitely liked how Pickover gathered interest in the problem through competition and storytelling.  As much as possible, I use both (in a politically correct way to avoid traumatizing my delicate students ) in my own teaching, and have found that helps in maintaining enthusiasm in the classroom.

Suppose that a square would have been fault-free except that there is a single tile that breaks the fault.  This means that the area on each side of the fault-line must have an odd area.  But we know that each tile has an even area (in fact, exactly 2 square units), so there must actually be an even number of tiles that break any fault line.

We also note that a single tile may only break one fault regardless of orientation, since the tile area is only 2 square units.  Now, when we consider a square, we see that there are 5 possible horizontal faults and 5 possible vertical faults.  This means that to construct a fault-free square, it is necessary to have at least 20 tiles (at least two per fault line), which gives a total area of 40 square units.  Unfortunately, squares have a total of 36 square units, so fault-free constructions are impossible.

Commentary:

This is from Arthur Engel’s Problem-Solving Strategies, 1997.  Engel says the original solution is from Golomb and Jewett, but then promptly doesn’t give a reference.  If anyone knows where the original solution is, I’d like to see it.  Engel also notes that we can extend this solution to show that a rectangle is fault-free if and only if is even, , and .  I haven’t actually done this part yet, but it seems plausible enough.  At some point in the next few weeks I’ll take a look at it.  If there are any surprises, I’ll let you know.

I like this problem (and its solution) partly because I have been enjoying tiling problems lately (I think it’s just a phase — my doctor assures me it won’t last) but mostly because this isn’t a contradiction proof, it’s just a failure-to-construct proof.  This always seems more positive and satisfying to me.  I suppose some may argue on semantics here .

I also really enjoyed what turned out to be a simple parity argument that I can easily explain to a class of low grade 9 and 10 math students.  And, the students find a potentially intractable problem to be a lot simpler than they thought, which is always reassuring for them in a classroom that they normally view as second only to jail.  Finally, despite the simplicity of the solution, it has left many of my bright grade 12 students completely stumped (me too initially, and my first version of a solution wasn’t nearly as clean as what I just showed you).

Consider a 6×6 rectangle, and cover it with 2 x 1 dominoes or tiles.  A fault-line is defined as a line from one side of the square to the other that is not also cutting through a tile.  Prove that any covering of the square must have at least one fault-line.

Problems and solutions will originate from such fine pedigree as geometry, number theory, pigeons and their domiciles, computer science, games and inequalities.  Of course, given the nature of math problems, especially fine ones, this list is certainly not exhaustive.

Now, given that every person is more than just the sum of his or her parts, and this seems to be a very one-sided part of me (math ‘n stuff), I suppose that on occasion, I will also wax eloquently about other things, although realistically, who knows where this will go?