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Posers and Puzzles

Posers and Puzzles

  1. 01 Aug '07 18:33
    There are 2 Series (S1 & S2)
    S1 begins at 10 and increases by 1
    S2 begins at 6 and decreases by 1

    Find the first whole number at which S1 is divisable by S2.

    *No Trial and Error allowed*
    10/6 = 1Mod4
    11/5 = 2Mod1
    12/4 = 3 Answer

    Just find how the answer 3(12/4) is resolved based on the initial inputs 10 & 6 and how you know each series behaves.
  2. 04 Aug '07 16:33
    We have to solve: (n+9)/(-n+7)=integer.

    But (n+9)/(-n+7)=-1+16/(-n+7).

    -1 is integer, so (n+9)/(-n+7)=integer iff 16/(-n+7) is integer.

    the biggest divisor of 16 smaller than 7 is 4. so -n+7=4, n=3, and the nubers are 12 and 4.
  3. 13 Aug '07 13:33
    The only problem with this is that once again you are you are using trial and error to find the first number less than seven divisable into 16. Can this be determined.

    Also,Can your initial work be applied to a more complex initial statement;

    102+n/21-2n - Once again finding the first n at which a whole number is produced. I'd presume 102+n is required to be multiplied by 2
  4. 14 Aug '07 15:35
    (n+102)/(-2n+21) = integer

    (-0.5)(-2n+21)+112.5/(-2n+21) = integer ... Multiply by 2

    2n-21+225/(-2n+21) = even integer

    225/(-2n+21) = odd integer

    225 = 15^2=3^2*5^2, so -2n+21=15, n=3
  5. Standard member uzless
    The So Fist
    14 Aug '07 17:04 / 1 edit
    Originally posted by jonnifandango
    There are 2 Series (S1 & S2)
    S1 begins at 10 and increases by 1
    S2 begins at 6 and decreases by 1

    Find the first whole number at which S1 is divisable by S2.

    *No Trial and Error allowed*
    10/6 = 1Mod4
    11/5 = 2Mod1
    12/4 = 3 Answer

    Just find how the answer 3(12/4) is resolved based on the initial inputs 10 & 6 and how you know each series behaves.
    highest common factor of 6 and 10 is 2.

    6 minus 2 is 4
    10 plus 2 is 12

    4/12 is the answer
  6. Standard member uzless
    The So Fist
    21 Aug '07 16:10
    no one debunked my solution?
  7. 23 Aug '07 16:03
    Uzless, i suppose that what jonnifandango wants is a formula to get the solution.

    I imagine the solution would be N="whatever", where N is the number you add to S1 and S2, and whatever will only depend on S1 and S2.

    I've started working it out with (10+n)/(6-n)=I , and i'm already stuck. I need to find out a proper definition for I.....
  8. 23 Aug '07 16:05
    I should have really started with (S1 + N)/(S2 -N)=I

    What can i replace "I" for?
  9. Standard member PBE6
    Bananarama
    23 Aug '07 17:23
    This is an example of a Diophantine equation, of the form:

    (a-bz) + n(1+z) = 0

    Where a, b are given and n, z are integers. I checked the Wolfram site and apparently there is a solution for linear Diophantine equations with 2 variables. I'll post the example when I get back from my meeting, but here's the link to the algorithm:

    http://mathworld.wolfram.com/DiophantineEquation.html
  10. Standard member PBE6
    Bananarama
    23 Aug '07 18:58
    Originally posted by PBE6
    This is an example of a Diophantine equation, of the form:

    (a-bz) + n(1+z) = 0

    Where a, b are given and n, z are integers. I checked the Wolfram site and apparently there is a solution for linear Diophantine equations with 2 variables. I'll post the example when I get back from my meeting, but here's the link to the algorithm:

    http://mathworld.wolfram.com/DiophantineEquation.html
    Oops, I guess this isn't really a linear equation. I'll try this little algorithm and see if it works anyway.
  11. Standard member uzless
    The So Fist
    23 Aug '07 19:06 / 1 edit
    Originally posted by yinyang78
    Uzless, i suppose that what jonnifandango wants is a formula to get the solution.

    I imagine the solution would be N="whatever", where N is the number you add to S1 and S2, and whatever will only depend on S1 and S2.

    I've started working it out with (10+n)/(6-n)=I , and i'm already stuck. I need to find out a proper definition for I.....
    Formulas are for the weak my yinyang!

    How's this:


    a+[factor a/b] =x
    b-[factor a/b] =y

    My formula is predicated on there actually being 2 numbers that will end up divided into each other