Fri. Dec 6th, 2019

Messy order | Science | THE COUNTRY


The solution to the "problem of the happy ending", raised last week, is 5. To see it more clearly, instead of stars consider 5 nails nailed halfway in a flat board: if we surround them with a loop and tighten it around the set until the string is taut, there are three possibilities: the string It 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, any 4 of the nails form a convex quadrilateral; in the second case, we already have the quadrilateral; and in the third, the 2 inner nails form a convex quadrilateral with 2 of the triangle. The attached illustration (taken from Clara Grima's article cited last week) shows it clearly.

Messy order



One would think that, if to have the certainty of being able to form a convex quadrilateral 5 points are enough, with 6 we will have the security of being able to form a convex pentagon; But this is not the case: a minimum of 9 points is required. And to have the certainty of being able to form a convex hexagon?

Primity and divisibility

Since the theme of last week was disorder, a concept as elusive as chance, many comments revolved around chaos, randomness and other issues that border on philosophy. And in that context our "outstanding user" Manuel Amorós proposed a curious problem:

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

The first of these huge 44-digit numbers would be: 19851986198719881989199019911992199319941995, which is obviously not a cousin because it ends in 5, and therefore is divisible by 5.

We can know 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 “outstanding user”, Francisco Montesinos, gave the correct answer, but without demonstrating it, so I invite my sagacious readers / is to mess up and reorder the eleven numbers from 1985 to 1995 so that they form a cousin, or show that they don't it's possible.

We can know 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 know immediately 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, to be divisible by 8 it must be the number formed by the last three figures.

Less obvious are the divisibility criteria by 3, by 9 and by 11. Do you remember them? Can you prove them? Can you deduce them?

Carlo Frabetti He is a writer and mathematician, a member of the New York Academy of Sciences. He has published more than 50 popular science works for adults, children and young people, includingDamn physics, damn mathorThe big game. He was a screenwriter ofThe Cristal ball.

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