14
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(Based on this puzzle by Gamow. The answer there might give you a hint (so don't look yet!))

At a fancy party in Café la Tour, everybody is friends with exactly 14 of the other people present.

  • Whenever two people at the party are friends, they have exactly 6 friends (present at the party) in common.

  • Whenever two people at the party are not friends, they have exactly 2 friends (present at the party) in common.

(Oh, and half of the people at the party came dressed in white, and the other half was dressed in black. How stylish!)

Question: How many persons are at the party?

(Note: Friendships are always mutual. Nobody is their own friend. There is at least one person at the party.)

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  • $\begingroup$ Why have you used chess ? $\endgroup$ – ghosts_in_the_code Nov 25 '15 at 14:36
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    $\begingroup$ Although the accepted anwer obviously is a solution, could there possibly be other solutions as well? $\endgroup$ – Ivo Beckers Nov 25 '15 at 15:04
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    $\begingroup$ @IvoBeckers: (hover spoiler) $\endgroup$ – Lynn Nov 25 '15 at 15:27
  • $\begingroup$ @Mauris thanks. actually the 4x4 still has one solution. It has different graphs yes but both have 16 vertices. $\endgroup$ – Ivo Beckers Nov 25 '15 at 16:15
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The chess tag gave it away.

There are exactly 64 people at the party.

To show this:

Imagine every person standing alone on a field on a big chessboard. 64 Fields and therefore 64 people.
Now every person is friends with those 14 other people which stand in the same row or column. (7 in the same row + 7 in the same column)

With this every two people which are friends and therefore are within the same row/column will have the other 6 people in the same row/column as common friends.

In a similar way every two people which are not friends and therefore neither within the same row nor the same column will have those two people as common friends that are standing on the intersections of the respective rows and column.

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  • $\begingroup$ Correct! (Did you spot the hint in the first sentence?) $\endgroup$ – Lynn Nov 25 '15 at 14:46
  • $\begingroup$ @Mauris Nope sorry. And I still can't seem to find the hint after looking again. $\endgroup$ – The Dark Truth Nov 25 '15 at 14:54
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    $\begingroup$ (hover spoiler). See also this link. $\endgroup$ – Lynn Nov 25 '15 at 14:59
  • $\begingroup$ @Mauris also the parenthetical, while it's most obvious use is to tell you there are an even number of guests, is, if not a hint, a confirmation afterwards, no? $\endgroup$ – Kate Gregory Nov 25 '15 at 15:53
  • $\begingroup$ @Kate: Yes, absolutely! I wanted to give lots of hints towards this solution, so that people unfamiliar with the math could still be inspired to try the solution first and then see it magically fall into place. $\endgroup$ – Lynn Nov 25 '15 at 15:57

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