Second day (1/7/00)

Today began with some informal conversation with the students about their interests, Found out that one of the students is participating in the Community Teaching Fellowship for Mathematics and Science Program and may want to be a teacher. Getting to know some names though quite a few are familiar after the first class because four were in my Math 100 class two years ago. One student I have known for a long time. I am also getting a sense of the backgrounds of these students and there appears to be a wide spread in their preparation. Before we get going I ask each student to put their name on a sheet of paper and record all the mathematics courses that have taken so far. I will use this as a guide to making the groups. In this way I can organize them by relative level of knowledge and sophistication so that there is less likelihood that one person might dominate a group because of their more advanced standing in the program. I will also try to balance the groups with men and women.

We got started by going over the problems from the previous class, beginning with the magic square problem. One student, representing her group, came to the board and simply wrote down a solution that was correct. When I asked if it was the only solution, she demurred. When I pointed out that one could rotate or reflect everyone recognized that there are, indeed, several solutions and together we figured out that there arise eight different ones, corresponding to each of the eight symmetries of the square. I asked if these are all of the solutions, which, because they come from anyone by such a symmetry, we can really think of as "one solution modulo symmetry." Another student came to the board and said that she proved the center square must be 5 is the average of the numbers 1, 2, . . . , 9. When I asked why it has to be the average, her explanation, "the center square is used the most" was not persuasive to the class. From there she said the magic number (the constant sum of the rows, columns and diagonals) must be 3 x 5 = 15. With these assumptions she was able to show that no corner can be odd. From there she was able to show that there are exactly the eight solutions found above.

A student, one of the more advanced ones, came to the board and used my hints from the previous time to show that the magic number must be 15 since the sum of three rows, for example, is three times the magic number and also equal to the sum of all the numbers, which is 45. However, he was unable to prove conclusively that the center was 5 (from which a complete determination of the problem would follow from the previous student's analysis).

At this point, my tendency to tell took over and I lead the class in finding the value of the center square by suggesting they add the two diagonal and the row and column which contain the center. This counts the magic number 4 times (and so sums to 60) and all the numbers once, except the center which is counted 4 times (three extra times). This gives an equation

60 = 45 + 3c (c for center) from which it follows that c = 5.

I then suggested another approach and asked the students if each square is in the same number of sums. The obvious answer was no: the center is in 4, the corners in 3 and the side elements in just 2. I asked them if, knowing the magic number is 15 could they use this to figure out the middle square, candidates for the corners and the sides. They took a few minutes and enumerated all sums of three integers from 1,2, . . . , 9 which add to 15 and saw that this did, indeed, lead to a complete determination.

We then discussed the locker problem. One student came to the board and reported on the result of some numerical experimentation which seemed to indicate that the squares were the open lockers. We then carried on a discussion of why this might be so. I posed some questions beginning with

What determines if a locker is open or closed at the end?

First answer, whether its state (opened when closed, closed when open) is changed an even number (closed) or an odd number (even) times. Without prodding a student said that the state is changed when the number of the person is a divisor of a locker number. This immediately led the class to conclude that the locker will be open if the number of its divisors is odd and closed if the number of its divisors is even. One of the advanced students then used prime factorization to show that precisely the even numbers have an odd number of divisors. One of the less experienced students had a more holistic and conceptual explanation: divisors of a number n come in pairs (a < b) unless n is a perfect square n = c^2 and then only one divisor arises from this factorization and so a square has an odd number of divisors.

I asked the students if they were all familiar with the pigeonhole principle. Most were, some were unsure and some certain they did not know it. I gave a brief, informal explanation which made sense to most of the students. I then pointed themto the lattice point problem and they worked in groups on this problem for the rest of class.