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32 people were invited at a party and started exchanging handshakes. Because of the confusion, each of them shook hands with each other multiple times: at least twice and up to X times. However, every two people exchanged different number of handshakes from every other two. What is the minimum possible number X, so that the above condition is met?

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  • $\begingroup$ OK all possible pairs are 32!/2!30! = 496, thus we must have at least 496 different numbers, starting from 2? $\endgroup$ – Samuel Apr 26 '16 at 9:10
  • $\begingroup$ So maybe the sum 2+3+...+496? $\endgroup$ – Samuel Apr 26 '16 at 9:21
  • $\begingroup$ The last sentence states the question clearly, whereas "the minimum number $X$ that each invitee shook hands with each of the others" is unclear and doesn't sound to me like what the last sentence is asking; I think you should rephrase that. $\endgroup$ – joriki Apr 26 '16 at 11:44
  • $\begingroup$ @joriki: Does it now make more sense? $\endgroup$ – Samuel Apr 26 '16 at 13:30
  • $\begingroup$ @Samuel: Yes, much clearer now. $\endgroup$ – joriki Apr 26 '16 at 14:12
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What am I missing?

I have clearly not understood the question as the answer for me is obvious. I would have liked to ask "what am I missing?" in a comment, but as I have less than 50 reputation I cannot yet make comments. So here is my understanding:

32 people meet and exchange handshakes with everyone else. Each pair shaking hands had between 2 and X handshakes. And no pair had the same number of handshakes. What is the minimum number of X?

With 32 people the number of pairs is, as you said, 496. So let one pair have 2 handshakes, another pair have 3 handshakes, another pair have 4 handshakes, etc, with the last pair having $2+496-1 = 497$ handshakes. Then X = 497.

So what am I missing?

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  • $\begingroup$ What if we change a bit the conditions, so that every invitee (as opposed to every pair) was engaged in a different number of handshakes? $\endgroup$ – Samuel Apr 30 '16 at 9:31

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