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Math Riddles & Trivia Here

  1. Sep 26, 2007 #1
    1. This thread is about mathematic riddles that we can all share, I will post one riddle and the next person list the solution and a new mathematic riddles as well (if possible), here is mine:
    If a stamp and an envelope cost a dollar and a dime (10 cents), and the stamp costs a dollar more than the envelope, how much does the envelope cost?

    2. Find the Solution

    3. Post a New Mathematic Riddle
  2. jcsd
  3. Sep 26, 2007 #2

    Math Is Hard

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    note: moved to Brain Teasers - MIH
  4. Sep 27, 2007 #3
    5 cents
  5. Oct 24, 2007 #4
  6. Nov 30, 2007 #5
    Try to count it. May be it will help you to accept that correct answer.
  7. Dec 2, 2007 #6
    How about, each can do 2 at the same time, so 6 simultaneous handshakes can happen at once?
    Otherwise that can do quite man, pretty much till they start to die off.
  8. Dec 13, 2007 #7
    15 is the right answer. n/2 * (n-1), where n is the number of participants. The -1 is to account for the participants not shaking their own hand.
  9. Dec 18, 2007 #8
    This is one-half of a 6x6 matrix with the diagonal center elements removed.
    Now, if they all spit at each other, how many spits are there?
  10. Feb 17, 2008 #9
    it's 30.
    there are 6 guys, and they each shake hands with 5 other people.
    so for the first guy; he shakes hands with 5 people, meaning there are 5 handshakes for that one guy.
    for the 2nd guy, there are another 5,
    for the 3rd there are another 5,
    and so on.
    so you do 6 x 5, which gives you 30.
    5 handshakes for each of the 6 men.

    butttt, i guess it can depend on whether or not a man can shake the same man's hand twice.
    then the answer would be less, but since it didn't include that in the riddle i won't include it while solving.
  11. Feb 18, 2008 #10
    Actually, in the original problem, "shaking hands with each other" usually means two people shake hands only once. The idea behind C(6,2) is that a handshake between person A and person B is the same as a handshake between person B and person A, and should be counted as only one handshake (hence the division by 2).

    Another way to think about it, going along with your line of thought, is that the first guy can shake hands with 5 different people. The second guy can shake hands with only 4 different people, because his handshake with the first guy was already counted. The third guy can only shake hands with 3 different people, because his handshake with the first and the second guys have already been counted, and so forth.

    Thus, we have not 5+5+5+5+5+5 = P(6,2) = 30 but 5+4+3+2+1+0 = C(6,2) = 15.

    Of course, if the problem really is a brain teaser, it could have some other senseless answer. Who knows?
  12. Mar 10, 2008 #11
    15 is the correct answer. 5 handshakes 4...3...2...1 = 15
  13. Jun 6, 2008 #12
    its 15
    6 people
    guy no 6 shakes hands with 5
    guy no 5 shakes hands with 4 [he's already done no.6]
    guy no 4 shakes hands with 3 [he's already done no.6, and no.5) etc etc
    ..........3 2
    ..........2 1

    thats 5+4+3+2+1 = 15

    whats the next puzzle?

    ok heres an oldie but a goodie

    how many people do you have to invite to a party such that:
    AT least 3 will be mutual friends
    OR at lest 3 will be mutual strangers?

    [mutual friends means A knows B and C, B knows C and A, C know A and B]
    same difference for mutual strangers!
  14. Jun 12, 2008 #13
    um i think you need more wording here... by what you gave i would say i could invite 3 people because 3 could be mutual friends OR they could just be 3 mutual strangers either way it could satisfy your problem based on your guidlines... did i miss something?
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