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Thread: 7 Math problems I just can't do

  1. #51
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    Quote Originally Posted by bi a op View Post
    17th I didn't understand this one, so I'll just translate what it says. Answers ABCD are included
    We're going to split a circle in as many pieces possible, with 7 straight lines.
    How many pieces can you get?
    14, 29, 35, 49, 128
    29.
    Basically each time you add another line, you add the previous amount of lines placed, but we get to start at 1.
    0 lines = 1 section
    1 line = 0 lines + 1 section = 2 sections
    2 lines = 1 line + 3 sections = 4 sections... etc
    3 lines = 4 +3 = 7 sections
    4 lines = 7 + 4 = 11 sections
    5 lines = 11 + 5 = 16 sections
    6 lines = 16 + 6 = 22 sections
    7 lines = 22 + 7 = 29 sections

    The pattern for this one also matches how you would do the 13th. 1, 2, 4, 7, 11, 16, 22, 29, 37...
    where f(x) is the number of sections with x lines, f(x) = (x*(x+1) +2)/2
    Last edited by Turpinator; 12-17-2012 at 04:17 PM.

  2. #52
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    agreed with turpinator 29
    Working on: Tithe Farmer

  3. #53
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  4. #54
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    New problem for the 18th!

  5. #55
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    #18:
    7 females.

    Simba Code:
    for f:=0 to 100 do
        for m:=0 to 100 do
          if (f + m) MOD 3 = 0 then
            if (13 * m + 10 * f + (1.0 / 3) * (m + f) * 6) = 159 then
              writeln(Point(f, m));

    Code:
    (7, 5)
    Last edited by riwu; 12-18-2012 at 01:09 PM.

  6. #56
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    Awkward moment when he only had 7 math problems for homework.
    You have all been played.
    Have you felt the whale lately?
    .__________.
    Whale so hard.

  7. #57
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    Luls. he wrote a script to 'bruteforce' the answer.
    Very nice though. Much easier than guess and check by hand eh?

  8. #58
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    Quote Originally Posted by Turpinator View Post
    Luls. he wrote a script to 'bruteforce' the answer.
    Very nice though. Much easier than guess and check by hand eh?
    Well there's no way to solve it algebraically since there's 3 unknowns but only 2 equations, so will have to do trial and error with f + m being multiples of 3
    The question is unconventional though, usually it should give us 3 equations for 3 unknowns then we can solve it with matrices (system of linear equations).

  9. #59
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    f * 10 + c * 6 + m * 13 = 159

    (m + f) / 3 = c



    f * 10 + ((m + f) / 3) * 6 + m * 13 = 159


    10f + 6((m + f) / 3) + 13m = 159


    10f + 2(m + f) + 13m = 159

    10f + 2m + 2f + 13m = 159


    12f + 15m = 159


    f = 159 - 15m / 12


    Since there can only be integers of f...


    m = 1,5,9


    f = 12,7,2


    *gasps* wow omg


    I believe the question posted isn't complete unless your teacher wants you to randomly pick 1..

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    It seems the solution is right...

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