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| Exactly 20 games (Posted on 2008-10-01) |
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A chessmaster who has 11 weeks to prepare for a tournament decides to play at least one game every day, but in order not to tire himself, he agrees to play not more than twelve games during any one week (consider "week" as periods of 7 consecutive days, starting from the first day of preparation; that is: if the first day is Sunday, each one of the eleven weeks ends on each one of the next eleven Saturdays.)
Prove that there exists a succession of days during which the master will have played exactly twenty games.
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Submitted by pcbouhid
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Rating: 3.2000 (5 votes)
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Solution:
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Starting with some initial day, say Monday, assume that the chess player plays a1 games; on Monday and Tuesday he plays a2 games; on Monday, Tuesday and Wednesday, he plays a3 games, and so on. By the end of the 77th day he has played a77 games.
Now, consider these two following sequence of integers:
a1, a2, a3, ..., a77;
a1+20, a2+20, a3+20,..., a77+20.
We have, in all, 2 x 77 = 154 integers, none of wich exceeds 132 + 20 = 152, inasmuch as a77 is not, according to the imposed conditions, to exceed 11 x 12 = 132 games played in eleven weeks. Hence at least two of these 154 integers must be equal.
However, no two integers of the first sequence can be equal, since the chess player has played at least one game every day; similarly, no two integers of the second sequence can be equal.
Therefore, we must have, for some j and some k, the equality:
aj = ak + 20.
This equation states that aj - ak = 20; so on a succession of (j - k) days - from the (j+1)st to the kth, inclusive - the chess master played exactly 20 games.
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