Pigeonhole in Chess Training

Solution

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Solution

Let a₁ be the number of games played on the first day, a₂ the total number of games played on the first and second days, a₃ 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 a₁, a₂, ..., a₇₇ is strictly increasing, that is, a₁ < a₂ < ... < a₇₇. Moreover, a₁ ≥ 1; and since at most 12 games are played during any one week, a₇₇ ≤ 12 × 11 = 132. Thus

Note that the sequence a₁ + 21, a₂ + 21, ..., a₇₇ + 21 is also strictly increasing, and

Now consider the 154 numbers

each of them is between 1 and 153. It follows that two of them must be equal. Since a₁, a₂, ..., a₇₇ are distinct and a₁ + 21, a₂ + 21, ..., a₇₇ + 21 are also distinct, then the two equal numbers must be of the forms a_i and a_j + 21. Since the number games played up to the i^th day is a_i = a_j + 21, we conclude that on the days j + 1, j + 2, ..., i the chess master played a total of 21 games.

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

Reference

1. Beifang Chen, The Pigeonhole Principle

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