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FoolForMath Group Title

Fool's problem of the day, How many ordered pair of \((x,y)\) are there such that \(x, y \in \mathbb{Z} \) and \( \frac 2 x - \frac 3 y = \frac 1 4 \) Good luck!

  • 2 years ago
  • 2 years ago

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  1. ParthKohli Group Title
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    Oops, maybe this involves some trial and error :P

    • 2 years ago
  2. ParthKohli Group Title
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    What is \(\mathbb{Z}\) in here?

    • 2 years ago
  3. FoolForMath Group Title
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    http://en.wikipedia.org/wiki/Integer

    • 2 years ago
  4. ParthKohli Group Title
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    This works with -8 and -6

    • 2 years ago
  5. ParthKohli Group Title
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    If b is -8 then a will be -16

    • 2 years ago
  6. ParthKohli Group Title
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    I see a lot of solutions to this :/

    • 2 years ago
  7. FoolForMath Group Title
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    Not many, less than 25 I believe ;)

    • 2 years ago
  8. ParthKohli Group Title
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    24 or 23 or 22.. I guess

    • 2 years ago
  9. ParthKohli Group Title
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    The minimum of b is I guess -36 maximum is 84

    • 2 years ago
  10. FoolForMath Group Title
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    Don't guess, try to form a analytic approach :)

    • 2 years ago
  11. ParthKohli Group Title
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    Oh wait, it goes further.

    • 2 years ago
  12. ParthKohli Group Title
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    Your problem is hard.

    • 2 years ago
  13. FoolForMath Group Title
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    No spoiler please :)

    • 2 years ago
  14. ParthKohli Group Title
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    How do we do it by the way?

    • 2 years ago
  15. FoolForMath Group Title
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    It's problem of the day, it's supposed to be interesting. However my approach takes less than 10 seconds ;) I will post it later though.

    • 2 years ago
  16. Arnab09 Group Title
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    is the answer 5? @FoolForMath ?

    • 2 years ago
  17. ParthKohli Group Title
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    No Arnab.

    • 2 years ago
  18. Arnab09 Group Title
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    okay, we can get after simplification.. x=8y/(y+12) as x is an integer, 8y must be a multiple of y+12 8y=2*2*2*y so, here are possible equations y+12=1 y+12=2 y+12=4 y+12=8 y+12=2y y+12=4y y+12=8y as y +12 is a factor of 8y all the equations except the last one will satisfy for an integral value of y so, the answer is.. 6 ordered pairs

    • 2 years ago
  19. Arnab09 Group Title
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    not 5^^

    • 2 years ago
  20. Arnab09 Group Title
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    sorry, there will be other options too if RHS is negative y+12=-1 y+12=-2 ................. and so on.. for y to be an integer, there are another 5 solutions so, total 11

    • 2 years ago
  21. rfig.khalil Group Title
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    8?

    • 2 years ago
  22. rfig.khalil Group Title
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    is it right?

    • 2 years ago
  23. ParthKohli Group Title
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    No, not right

    • 2 years ago
  24. rfig.khalil Group Title
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    the equation is equivalent to (x-8)*(y+12)=-96 so the answer is 11

    • 2 years ago
  25. ParthKohli Group Title
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    Na na

    • 2 years ago
  26. rfig.khalil Group Title
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    is all the way wrong or just the 11?

    • 2 years ago
  27. rfig.khalil Group Title
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    22

    • 2 years ago
  28. rfig.khalil Group Title
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    I m pretty sur it's 22

    • 2 years ago
  29. rfig.khalil Group Title
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    24

    • 2 years ago
  30. rfig.khalil Group Title
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    is 24 correct answer.?

    • 2 years ago
  31. ParthKohli Group Title
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    No, you are close by the way

    • 2 years ago
  32. joemath314159 Group Title
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    I was able to find 11 ordered pairs that are solutions: (7, 84) (6,36) (5,20) (4,12) (2,4) (-4,-4) (-8,-6) (-16,-8) (-24,-9) (-40,-10) (-88,-11) I think there might be more, but im missing an idea.

    • 2 years ago
  33. FoolForMath Group Title
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    HINT: There are 23 of them.

    • 2 years ago
  34. FoolForMath Group Title
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    Real challenge is to find a quick analytic approach :)

    • 2 years ago
  35. joemath314159 Group Title
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    yeah i just found the other 12, making 23.

    • 2 years ago
  36. joemath314159 Group Title
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    Most people probably got that:\[8y-12x-xy = 0\]There is a way you can factor this to get an idea of how many solutions there are...

    • 2 years ago
  37. FoolForMath Group Title
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    That's right joe! Congrats man!

    • 2 years ago
  38. joemath314159 Group Title
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    ty :)

    • 2 years ago
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