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Challenging Chess Championship (Posted on 2005-11-16) Difficulty: 3 of 5
Eight players competed at a recent chess tournament. Knowing that:

  • Each player played all the others, exactly once.
  • Winning earns you 1 point; drawing, ½ point; and losing, 0 points.
  • Everybody ended with a different number of points.
  • The one who ended 2nd earned as many points as the four bottom players put together.
  • What was the result of the game between the player who ended 3rd and the player who ended 7th?

    See The Solution Submitted by Old Original Oskar!    
    Rating: 3.3333 (6 votes)

    Comments: ( Back to comment list | You must be logged in to post comments.)
    Some Thoughts starting at the bottom | Comment 9 of 13 |
    Consider the bottom two scores. They can't be 0 and 0 since they must be distinct. They also can't be 0 and 0.5 because the 0 lost to everyone and the 0.5 has no corrsponding win.

    Also, if anyone scored 7, then nobody scored 6.5 since everyone has at least one full point  lost to the 7. Then the 2nd highest is at best 6. If the highest scorer is 6.5 then the 2nd highest is again maximally 6.

    since 0,0.5 and 0,1 are not possible lowest scores, the smallest sum of the lowest four scores is at least 6 (0,1.5,2.2.5; 0, 1, 2, 3; or  .5, 1, 1.5, 2). Since the sum of the bottom fout is at least 6 and the 2nd place score is at most 6 and the two are equal, they're both 6. And that means the bottom scores are one of the three possibilities listed. (Aside: 0,1,1.5,* is impossible by the same argument against 0,0.5 -- the 1 lost to all but the zero so the next highest person has at least 2 wins. Similarly 0,1,2,2.5 is not possible.)

    Now consider each possibility for the four lowest scores. In the first, the 1.5 score beat the 0 and tied one other. The 2 therefore beat the 1.5 and the 0, and the 2.5 could not have beat the 2, 1.5 AND 0 and so must have beat the 2 and the 0 and tied the 1.5. Since all of these players wins are accounted for, they lost to everyone else. Specifically, #3 beat #7. In the second case, the 1 beat only the zero, the 2 beat the 1 and 0, and the 3 beat the 0,1, and 2 and again all four lost to the rest, and #3 beat #7. The third possibility is impossible, for it requires that the 1 beat the 0.5, the 1.5 then must have beat the 1 and tied the 0.5, and that forces the "2" to have won three games against the 0.5, 1, and 1.5. (Alternately, the 1 could have tied the 0.5 and another player, but then the 1.5 would have had to beat the 0.5 and tie the 1 and again the 2 would have had to beat all three of the previous.)

    It may not be possible to determine the actual scores of all eight players, but the #2 player scored six points and the #3 player beat the #7 player since all of players 1-4 beat players 5-8.


      Posted by Paul on 2005-11-17 02:34:14
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