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A Handshake Problem

  1. Apr 15, 2007 #1
    Brian and Steve were invited to a party attended by four other pairs of siblings. So, a total of ten persons were present at the party. During the party various handshakes took place, but no person shook their own hand or the hand of their sibling.

    At the end of the party Steve asked each person, including Brian, how many different people they shook hands with, and was surprised to note that every number was different.

    How many hands did Brian shake?
  2. jcsd
  3. Apr 15, 2007 #2
    At a first glance it seems to me as there would be multiple possible answers. I'll have to think about this some more.

    Are you sure all needed information is provided?
  4. Apr 15, 2007 #3


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    Seems to be a bit of a problem, here. If nobody shook their siblings' hands, the highest number of different people that any person could share a handshake with is 8 and the smallest number is 0. If each of the 10 people shook hands with a different number of people, the numbers 0-8 would not allow each of them to have a unique number of handshakes.
  5. Apr 15, 2007 #4
    0-8 means that there are 9 options. Steve was asking the 9 others, that does not imply he is part of the set of unique handshakes.
  6. Apr 15, 2007 #5


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    True - the problem statement did not say that Steve had to consider himself - only the people that he asked. Thank you.

    Edit: Interesting - Brian cannot have shaken 0 hands because if anybody other than Steve (who is not in the 0-8 grouping) had shaken 8 hands, he would have had to have shaken Brian's hand, as well as Steve's. Since Brian cannot have shaken 0 hands if anybody else shook 8 hands, nobody else is in a position to have shaken 8 hands except Brian. He was the high-scorer.
    Last edited: Apr 15, 2007
  7. Apr 16, 2007 #6
    One thing isn't clear to me-- is Brian Steve's sibling? It doesn't really state this, but I'm not sure if I should assume it to be true. If not, then the possibilities are 1-9 or 0-8. Essentially, the same logic applies, meaning that either way, Brian shook 8 or 9 hands. 8 if they're siblings, and 8 or 9 if they're not. Maybe that's the crux of the problem right there-- assume that they're *not* siblings, and now you actually have to figure out a solution where the shaking works out correctly!


    [edit]Ok, wrong assumptions. I assume that Brian and Steve ARE siblings, making the answer unique. And no, it's not 8. If they AREN'T siblings, then there are multiple solutions, as asserted above. ... Further if they AREN'T siblings, Brian has 6 possibilities for the number of handshakes that he could've had.[/edit]
    Last edited: Apr 17, 2007
  8. Apr 16, 2007 #7
    I still don't think I'm understanding the logic needed to solve this problem.

    I understand that the number of handshakes possible are 0-8. Each one is assigned to each of the people (besides Steve).

    The part that gets me is that I don't see ANY distinguishing traits about Brian has compared to all of the other people at the party.
  9. Apr 17, 2007 #8
    Try actually drawing out a solution. Hint:
    Start with whoever shakes 8 hands. You'll notice that the sibling of whoever shakes 8 hands *has* to be the one who shakes 0 hands. Next, arbitrarily choose someone else at the party who shakes 7 hands. Who does his sibling have to be? Etc.
  10. Jan 19, 2008 #9
    Under this conditions the task is a paradox. There will be 3 people minimum with the same amount of handshake - 4. Then the author should rephrase it that he didn't count himself and his sibling. Then his sibling just like himself had 4 handshakes. And yes draw the table 10x10 and match their handshakes giving the #1 person 8 handshakes, #2 - 7 handshakes and so on. Soon you'll see that when you'll reach 4 handshakes they will repeat in 2 more persons and only then will drop to 3, then to 2 and to 1. With all other variations you'll get even more same numbers of handshakes.
    Am I right?
  11. Jan 19, 2008 #10
    Well, no. First of all, there are a minimum of 2 out of 10 people who shook the same number of hands at the party, not 3. But beyond that, the problem is correct because it doesn't say that Steve didn't shake the same number of hands as everyone else, it just says that none of the people Steve asked shook the same number of hands. Hence, someone shook the same number of hands as Steve. That's actually pretty obvious, since the maximum number possible would be 9, and the lowest number possible would be 0, and we know for a fact that both 9 and 0 cannot co-exist in this example. Hence, the range must be 1-9 or 0-8, which each have 9 numbers in them. And since there are 10 people, it's a verifiable fact that at LEAST two of the people shook the same number of hands. It's just that one of those people happens to be Steve, who's doing the asking.

  12. Jan 19, 2008 #11

    My original answer was "4", but PF system did not accept it, because at least 4 characters needed, That is why I posted this PS. :smile:

    Steve got answers 0 to 8. Somebody said 8. So, there are 8 + 1 = 9 persons, who shook hands at least 1 time. But who shook hands 0 times? There is only one answer. Sibling of a person, who said 8. Thus, we have numbers of handshakes grouped by siblings.

    7-1 (somebody said 7, who said 1? His/her sibling) etc.

    There is only one way for Steve to get 9 different answers. He shook hands 4 times and so did his sibling Brian.
    Last edited: Jan 19, 2008
  13. Jan 19, 2008 #12
    The only possible way i can see this could be true is that some of them can't count, but I haven't read replies yet, so explanations may be there to help...
  14. Jan 20, 2008 #13
    It's too hard to determine because I know that not everyone likes to shake hands. Various people shaking hands does not mean that everyone shook hands. I am reluctant to shake hands with just anyone as some people have sweaty palms. I know its fussy but you have to take things like that into account.
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