## Teaser 1986: Protracted calculation

**From The Sunday Times, 8th October 2000** [link]

In the circle illustrated the numbers represent the sizes of the sectors:

By combining adjacent ones it is possible to find sectors in this circle of sizes 1, 2, 3, … all the way up to 13 (13 being the whole circle). For example:

7 = 4 + 1 + 2.

In a similar fashion, given a circle divided into five sectors in a particular way, it is possible to combine adjacent sectors to give sizes 1, 2, 3, … up to the biggest possible in the circumstances.

What, in increasing order, are the sizes of the five sectors?

The text of this puzzle is taken from the book *Brainteasers* (2002, edited by Victor Bryant), the wording differs only slightly from the puzzle originally published in the newspaper.

[teaser1986]

## Jim Randell 7:09 am

on21 March 2019 Permalink |If the circle is divided into

ksectors, then for each starting sector (and there arekof them) we can make a contiguous region consisting of 1, 2, 3, …,(k – 1)sectors. We don’t make a region ofksectors because that is the whole circle, so we add that in separately giving a total number of arrangements of:To make the arrangements correspond to the numbers 1, 2, 3, …

n(k), the whole circle needs to correspond to the valuen(k), and there needs to be a single sector corresponding to the number 1. So we place the number 1 in the first sector, and then distribute the remaining(n(k) – 1)between the remaining(k – 1)sectors.This Python 3 program finds the solution in 83ms.

Run:[ @repl.it ]Solution:The five sectors have values: 1, 2, 3, 5, 10 (in numerical order).They can be arranged like this:

As well as the example given it is also possible to make a circle with 4 sectors using the values: 1, 3, 2, 7.

The program is suitable for experimenting with small values of

k. (One simple way to improve the program is to note that as well as a 1 sector, we will also need a 2 sector in the remaining decomposition).Here are the number of solutions for various values of

k:(See: OEIS A058241 [ link ]).

We’ve come across the following formula before:

Specifically in the exploration of

Teaser 2907, where it is the number of elements in afinite projective planeof order(k – 1).The fact that there is no solution for

(k = 7, n = 43),(k = 11, n = 111)and(k = 13, n = 157)makes me wonder if there is a link with projective planes, as finite projective planes of order 6, 10, and 12 (probably) do not exist.For a more efficient way to generate “magic circles” see my comment on

Enigma 985.LikeLike