#2             85. A REGULAR PROBLEM             
You are given a box of matches. With these matches you find that you can form all possible pairs of the four regular figures: triangle, square, pentagon and hexagon, using all the matches every time.

For example, if the box contains 11 matches, you could form the three pairs of figues illustrated below, but not the triangle and the hexagon, or the square and pentagon, or the square and hexagon. Hence the box cannot contain 11 matches.

Of course no polygon can have sides of varying lengths.

What is the smallest number of matches in the box?
HINT 1

Get a box of matches and play! Try out various combinations to see what happens.
HINT 2

Do you think the solution may have something to do with divisibility? For example, for a pentagon, what numbers would be possible?
SOLUTION

Let n be the number of matches. Now T + H shows that n must be a multiple of 3. S + H shows n must be a multiple of 2. Hence n is a multiple of 6. P + H shows that n is of the form 5p + 6h. But since n is a multiple of 6, this must be 30p' + 6h. So n is at least 36. You can check that n = 36 will do.
EXTENSIONS

1. Try varying the number of polygons. What happens for one polygon? Three?

2. You could try a different selection of polygons. Even including a square and a rhombus might add interest (and test the puzzler’s knowedge!).