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This question already has an answer here:

At a conference there are 2016 participants, numbered 1 to 2016. Each participant 1 through 2015 shook hands with as many people as his number. How many hands did the 2016th participant shake?

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marked as duplicate by Apep, Quintec, Rand al'Thor, ffao, Rubio Mar 15 '18 at 22:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ In that question it is possible for someone to have 0 friends. $\endgroup$ – Tmpod Mar 15 '18 at 21:33
  • $\begingroup$ Is it possible for someone to have 0 hands? A conference of amputees, maybe? :) $\endgroup$ – Chowzen Mar 15 '18 at 21:40
  • $\begingroup$ The solution to this puzzle follows from the solution to the {0,1...13} situation in the other puzzle; you just have to ignore the first person. $\endgroup$ – Apep Mar 15 '18 at 21:41
  • $\begingroup$ Welcome to Puzzling SE, take a tour here puzzling.stackexchange.com/tour to earn a badge! :D $\endgroup$ – QuantumTwinkie Mar 15 '18 at 22:33
  • $\begingroup$ I believe this leads to a different method of solution and significantly different answer than the 'Tiffany' question. So while @Apep correctly points out that 'you just have to ignore the first person', it's not immediately obvious how that will affect the answer. I won't put my answer here because I can't spoiler tag it. $\endgroup$ – theonetruepath Mar 16 '18 at 0:14
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Well I didn't look at the possibly duplicate one and here's an answer:

2015 shakes hands with everyone (because there's only 2105 other people)
1 only shakes hands with 2015 (because that's one person already)
2014 shakes hands with everyone except 1 (only that many people left)
2 only shakes hands with 2015 and 2014 (that's two people already)
... and so on
1009 shakes hands with everyone from 1007 upwards
1007 shakes hands with everyone from 1009 to 2015 (inclusive)
1008 shakes hands with everyone from 1009 to 2016
So, 2016 shakes 1008 people's hands (specifically the people numbered 1008 to 2015)

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  • 1
    $\begingroup$ Yes, this is correct! $\endgroup$ – Tmpod Mar 18 '18 at 18:01

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