Provided one of the integers m and n is even, it is obvious that an m × n rectangle can be covered with 1/2mn dominoes, each domino covering two unit squares. If some of the dominoes are oriented in each of the two possible ways, it is possible that there may be no continuous straight line of domino edges, either horizontal or vertical, across the rectangle: in other words, each of the (m — 1) + (n — 1) fault lines on the underlying grid is crossed by at least one domino. Such rectangles are called fault-free, and are mentioned by S. Golomb in [1] where the concept is attributed to R. I. Jewett. An example with m = 6, n = 8 is shown in Fig. 1.

An elementary algebraic problem attributed to Leonardo of Pisa is analysed and some illogical elements in its formulation and solution are exposed. The natural context in which the problem was formulated is then proposed, and some consequences are discussed.

How many times have you heard that somebody is a wizard? And how many times did you take it literally? Most likely, the answer to the second question is ‘never’. And yet, there are reasons to believe that some people among us are real wizards, of the kind described with so much charm in the recently published series of books on Harry Potter [ 1, 2, 3, 4 ]. These books have provided us with a wealth of details on wizards and their society.

A ball is shot at an angle of 45 degrees from the corner of a rectangular billiard table of size m × n (where m and n are integers). It always bounces off from a side at 45 degrees until it lands in a corner pocket. Will it always land in a corner or will it keep bouncing for ever? And how many times will it bounce? And which corner pocket will it land in?

Try it for yourself. For example in the 7 × 4 table shown the ball will bounce 9 times and then fall into the top left-hand corner pocket.

This is a well-known problem but we now present a simpler proof than we've seen elsewhere of the fact that the ball always ends up in a corner pocket, and we derive a formula for the number of bounces and we determine the finishing pocket.

How many ways are there of making up £1 using coins? Omitting the use of the £1 coin, we are asking for the number of ways of ‘partitioning’ 100 using parts of 1, 2, 5, 10, 20, 50. For n ≥ 0, let f (n) denote the number of solutions to the equation

in non-negative integers xi, so that our required number is f (100). It is not difficult to see that f(n) is the coefficient of tn in the product Π(1 + tk + t2k +… ) over the parts k = 1, 2, 5, 10, 20, 50.

The Fifteen puzzle is well known. In its original form it consists of a shallow square box that can just hold sixteen small squares. Squares, numbered 1 to 15, are placed in any order in the box, leaving the sixteenth position empty (e.g. Fig. 1). The problem is to slide the squares, without lifting them out, until the numbers are in their natural order (Fig. 2).