# Pigeonhole in Chess Training

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny### Solution

Let a_{1} be the number of games played on the first day, a_{2} the total number of games played on the first and second days, a_{3} the total number games played on the first, second, and third days, and so on. Since at least one game is played each day, the sequence of numbers a_{1}, a_{2}, ..., a_{77} is strictly increasing, that is, a_{1} < a_{2} < ... < a_{77}. Moreover, _{1} ≥ 1;_{77} ≤ 12 × 11 = 132.

_{1}< a

_{2}< ... < a

_{77}≤ 132:

Note that the sequence a_{1} + 21, a_{2} + 21, ..., a_{77} + 21 is also strictly increasing, and

_{1}+ 21 < a

_{2}+ 21 < ... < a

_{77}+ 21 ≤ 132 + 21 = 153.

Now consider the 154 numbers

_{1}, a

_{2}, ..., a

_{77}, a

_{1}+ 21, a

_{2}+ 21, ..., a

_{77}+ 21.

each of them is between 1 and 153. It follows that two of them must be equal. Since a_{1}, a_{2}, ..., a_{77} are distinct and _{1} + 21,_{2} + 21,_{77} + 2_{i} and _{j} + 21.^{th} day is _{i} = a_{j} + 21,

(This problem is reminiscent of couting aspirins taken taken on consecutive days.)

### Reference

- Beifang Chen, The Pigeonhole Principle

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2008 Alexander Bogomolny64956442 |