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Maths Handshake Puzzle

Hands shakeMaths Handshake Puzzle - 4 September

At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?

For Solution : Click Here

25 comments:

  1. 12 People:

    #12 shakes w/ #1-11 (11 shakes)
    #11 shakes w/ #1-10 (10 shakes)
    #10 shakes w/ #1-9 (9 shakes)...

    11+10+9+8+7+6+5+4+3+2+1+0=66 total handshakes

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    Replies
    1. correct nC2 = 66 , and n comes out to be 12.
      n*(n-1) = 66*2 and 12 satisfies this relation

      Delete
  2. ans: 13 members
    for ex: if 4 mems are there then 6 shakes... i.e. (n-1*(n))/2 so here n-1*n/2=66, then n is 13

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  3. This comment has been removed by the author.

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  4. ((13-1)*13))/2 = 78
    ((12-1)*12))/2 = 66

    Still going with 12 :)

    Also, check the parentheses: ((n-1)(n))/2

    As written, (n-1*(n))/2 simplifies to (n-n)/2, and thence to, well, 0.

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  5. nc2= n*(n-1)/2=66
    n=12
    the answer is 12 people.

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  6. 1+2+3+....+(n-1) = 66; total persons = n
    n(n-1)/2=66 => n(n-1) = 132 => n = 12
    Explaination: If there are n people, the first person
    will shake hand with n-1 persons, the second with n-2, the third with n-3, and the (n-1)th person with n-(n-1)=1
    person; ie, the last person. Hence, total handshakes are (n-1)+(n-2)+(n-3)+....+3+2+1

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  7. 12 people....
    coz,

    n = total persons
    totally 66 handshakes,
    2 persons = 1 handshake
    and each person shakes hand with n - 1 persons

    ==> Half of ( n(n-1) ) = 66
    ==> n(n-1)/2 = 66
    ==> n = 12 or -11

    ==> n = 12

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  8. 12, it's so simple

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  9. Here's the script i made, just save as anything.bat, 14 here, :
    @echo off
    set people=1
    set handshakes=0
    set cnt=0
    :LOOP
    set /A people+=1
    set /A cnt+=1
    set /A handshakes+=%cnt%
    echo.handshakes:%handshakes%
    if "%handshakes%" NEQ "66" Goto :LOOP
    echo.Handshakes:%handshakes%
    echo.People:%people%
    pause > nul
    exit /b

    ReplyDelete
  10. 12

    In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n.
    Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 66.
    This is the quadratic equation n2+ n -132 = 0. Solving for n, we obtain 11 as the answer and deduce that there were 12 people at the party.

    Since 66 is a relatively small number, you can also solve this problem with a hand calculator. Add 1 + 2 = + 3 = +... etc. until the total is 66. The last number that you entered (11) is n.

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  11. it's just nC2=66
    so n=12

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  12. This comment has been removed by the author.

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  13. (X-1)+(X-2)+(X-3)+......+{X-(X-1)}=66
    So X=12

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  14. It's basically 11+10+9+8+7+6+5+4+3+2+1+0 which equals 66
    The answer is 12. Cmon people it was so easy.



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  15. In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n.
    Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 66.
    This is the quadratic equation n2+ n -132 = 0. Solving for n, we obtain 11 as the answer and deduce that there were 12 people at the party.

    so the answer is 12.

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  16. 12 people. Number of shakes=n*(n-1) where n is number of people

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  17. 33 because 33x2=66

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