## Teaser 3139: Checkmate

**From The Sunday Times, 20th November 2022** [link] [link]

Our chess club is divided into sections, with each section having the same number of players. The two oldest members will soon be retiring from playing and we will change the number of sections. The number of sections will change by one and so will the number of players in each section, but all the sections will still have the same number of players. This will result in there being a grand total of 63 fewer matches per year if each member plays all the other members of their section once per year.

How many players are in each section at the moment?

[teaser3139]

## Jim Randell 4:56 pm

on18 November 2022 Permalink |If the players are split into

ksections, each consisting ofnplayers, then the number of matches in each section isC(n, 2), and the total number of matches ism = k.C(n, 2).We can use some analysis to simplify the puzzle to two cases, one of which has no solution for positive integers, and the other that gives us the required solution.

However, here is a constructive Python program that considers the total numbers of players

t(up to 2000), and generates the possible(n, k, m)values for each candidatetvalue. It then looks for twotvalues that differ by 2, and givenandkvalues that differ by 1, andmvalues that differ by 63.This Python program runs in 85ms.

Run:[ @replit ]Solution:There are 10 players in each section.The club starts with 110 players, formed into 11 sections of 10 players each. There are 45 matches among the players in each section, and 495 matches in total.

When 2 players leave there are 108 players remaining. These are formed into 12 sections of 9 players each. There are 36 matches among the players in each section, and 432 matches in total. This is 63 fewer matches.

Manually from:

we can see that (for

n+k> 3) there are 2 cases:[Case 1](n, k)→(n + 1, k − 1)⇒k=n− 1.Which has no solution in positive integers.

[Case 2](n, k)→(n − 1, k + 1)⇒k=n+ 1.And so there are 11 sections, each with 10 players.

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## GeoffR 3:49 pm

on19 November 2022 Permalink |LikeLike

## GeoffR 6:21 pm

on19 November 2022 Permalink |LikeLike