The solution to the "happy ending problem," raised last week, is 5. To see it more clearly, instead of stars let's consider 5 half-nailed nails on a flat board: if we surround them with a loop and squeeze it around the set until the string is tense, there are three possibilities: the string forms a pentagon with the 5 nails as vertices, the string forms a quadrilateral with 4 nails as vertices and 1 inside, the string forms a triangle with 3 of the nails as vertices and 2 inside. In the first case, 4 of the nails form a convex quadrilateral; in the second case, we already have the ring; and in the third, the 2 inner nails form a convex quadrilateral with 2 of those of the triangle. The accompanying illustration (taken from Clara Grima's article quoted last week) shows this clearly.

ittly be thought that, if to be sure that we can form a convex quadrilateral 5 points are enough, with 6 we will be sure to be able to form a convex pentagon; but it is not so: for this it takes a minimum of 9 points

.### Primity and divisibility

Since last week's theme was disorder, a concept as elusive as the random one, many comments revolved around chaos, randomness and other issues that bordered on philosophy. And in that context our "featured user" Manuel Amorós proposed a curious problem:

If we have the integers between 1985 and 1995 (both included) followed in a certain order, can the number obtained in any case be a prime number?

The first of these huge 44-digit numbers would be: 1985198619871988198919901991199219319941995, which is obviously not prime because it ends in 5, and is therefore divisible by 5.

We can tell if a number, however long, is divisible by 5 without more than seeing its last figure: if it is a 0 or a 5, the number will be divisible by 5; otherwise it will not be

Another "standout user", Francisco Montesinos, gave the correct answer, but without demonstrating it, so that I invite my shrewd readers to mess up and reorder the eleven numbers from 1985 to 1995 so that they form a cousin, or to prove that it is not possible.

We can tell if a number, however long, is divisible by 5 without more than seeing its last figure: if it is a 0 or a 5, the number will be divisible by 5; otherwise it will not be. We can also immediately know if it is divisible by 4, because for this the number formed by its last two figures must be a multiple of 4. And, similarly, for it to be divisible by 8 it has to be the number formed by the last three figures,

Less obvious are the criteria of divisibility by 3, by 9 and by 11. Do you remember them

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