Home > Riddles > The party without different grips
At a party there are $n$ people, with $n\ge 2$. Each person writes down how many handshakes they gave during the party.
Question: Can $n$ all different numbers appear between those counts? That is, is it possible that one person has given 0 squeezes, another 1, another 2, ..., and another $n-1$?
Back to problem
Answer: No, it is impossible.
If there were $n$ counts that were all different between $n$ people, they would necessarily be:
$$ 0,1,2,\dots,n-1. $$
But then they would coexist:
The one from $n-1$ had to greet everyone, including the one from 0. Contradiction.
Therefore, in any party with $n\ge2$, at least two people share the number of squeezes.
If you enjoyed this one, try more pure-logic riddles, explore this theme, browse the full archive, or read the riddle-solving guide.