riddle of the day

The scene is social, but the problem has a more rigid structure than it seems. It is worth asking what restrictions the distribution of squeezes actually imposes before answering.

The squeeze party

Reasoner
Pure logic

Riddle statement

At a party there are \(n\) people, with \(n\ge2\).

Each person writes down how many handshakes they gave during the party.

Is it possible that the \(n\) numbers recorded are all different?

Show solution

Solution

Answer: No, it is impossible.

If the \(n\) counts were all different, they would have to be exactly:

$0,1,2,\dots,n-1.$

That would force them to coexist:

  • a person with 0 grips;

  • a person with \(n-1\) squeezes.

The one from \(n-1\) had to greet everyone, including the one from 0. Contradiction.

Therefore, at any party with \(n\ge2\), there are always at least two people who end up with the same number of squeezes.