Pussy Beware!

Discussion in 'Math' started by studiot, Mar 3, 2009.

  1. studiot

    Thread Starter AAC Fanatic!

    Nov 9, 2007
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    They've counted the cats in Llanfair
    Which number a third of a square
    If a quarter were slain
    Just a cube would remain
    How many, at least, must be there?


    Old but good question.
     
  2. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Thought this out, don't ask me for an equation. :eek:

    108 (assuming no partial cats were slain)
     
  3. wr8y

    Active Member

    Sep 16, 2008
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    They've counted the cats in Llanfair
    Which number a third of a square

    108 is a 1/3 of 324.
    324 is 18 squared - ok so far.

    If a quarter were slain
    Just a cube would remain


    If a quarter were slain, that would be 3/4 times 108 = 81.

    Just a cube would be there

    81 is 4.3267487 to the 3rd power.

    This is not a "cube", 2 cubed is 8 but this is just a number cubed.
    I think you are wrong - or am I confused?
     
  4. jpanhalt

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    Jan 18, 2008
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    I got 972 cats.

    John
     
  5. Ratch

    New Member

    Mar 20, 2007
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    To the Ineffable All,

    The answer is 972 cats. Here's why. You have to find out what number makes a perfect cube root of 3x/4. So x has to complete the cube and be divisible by 4. The first number that does that is 36. So three quarters of 36 is 27, which is a perfect cube. The second test is whether 3 times the number is a perfect square. Three times 36 is 108 which is not a perfect square, so the number 36 fails.

    The next number that completes the cube and is divisible by 4 is 3x9x9x4 = 972 . Three quarters of 972 is 729, which is a perfect cube of 9. Three times 972 is 2916 which is a perfect square of 54. That proves that 972 is the smallest number that satisifies the conditions of the problem.

    Ratch
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    I am wondering if I misread the verse.

    First Condition: They've counted the cats in Llanfair
    Which number a third of a square

    NumberOfCats\Rightarrow  \sqrt[]{\frac{Perfect Square}{3}}

    Second Condition: If a quarter were slain
    Just a cube would remain
    PerfectCube=\sqrt[3]{\frac{NumberOfCats}{4}}

    Modifier: How many, at least, must be there?

    To me, this implies the first two aren't sequential, but two statements, such as:

    "Joe is Taller than Jill, Mary is Taller than Todd. Name them in order of height"

    Rather than finding a condition where two of the sentences are true.


    --ETA: On The Other hand, If you are a Sadistic type, your answer is correct. :D You musta been the kid that kept running up in class to look inside Schrödinger's box.


    --ETA AGAIN! OK, I WAS WRONG! I Forgot the first sentence already whacked most the kitties.
     
    Last edited: Mar 3, 2009
  7. studiot

    Thread Starter AAC Fanatic!

    Nov 9, 2007
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    You folks were quicker off the mark with this one.

    Well done Ratch

    Edit: sorry and John.
     
    Last edited: Mar 4, 2009
  8. fx12

    New Member

    Mar 4, 2009
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  9. studiot

    Thread Starter AAC Fanatic!

    Nov 9, 2007
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    Welcome to AAC, FX12

    I didn't claim originality and I didn't find it on the internet. The ditty is much much older than the net.
     
  10. thatoneguy

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    Feb 19, 2009
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    You used google, and didn't post before Ratch (or anybody... Welcome!)

    I proudly displayed my idiocy in public. It was fun writing in TeX though. :)

    So you have a plane on a treadmill.....
     
  11. fx12

    New Member

    Mar 4, 2009
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    thatoneguy "idiocy..." not at all you made a real attempt and came close.
    If I were to say that you were the one to come closest to the answer-because studiot, Ratch (not saying they did but it is possible) and I found the answer on the net-would you still say "idiocy..." My point was to show that we can not compare ones knowledge, IQ etc because we are all on the net?
     
  12. jpanhalt

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    Jan 18, 2008
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    Ratch,
    Thank you for explaining my answer, and as usual, you are right.

    The way I reached that answer: x= # of cats, a and b are variables. x, a, and b are integers.

    1) x = 1/3a^2
    2) 3/4x = b^3
    3) 4b^3 = a^2

    I found solutions to equation #3 and tested them in equation #1. One can rule out certain values for "b" immediately by inspection and logic, like 4. I didn't consider primes for "b" because of equation #2. 8 was too similar to 4. It just seemed that "b" might be 9 or 12; although, I did try 6 too.

    Once 9 was found to be viable for "b", the rest of the problem was solved.

    John
     
  13. b.shahvir

    Active Member

    Jan 6, 2009
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    can someone call the PETA guys pleazzzz ? :D
     
  14. wr8y

    Active Member

    Sep 16, 2008
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    After getting a "B" in College Algebra and an "A" in Precalculus last year (working on an ASEET), I hang my head in shame for not being able to do that.

    Thanks! That was a nice flashback to Mr. Lee's classes....
     
  15. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    True.

    I'm also a newbie, but have seen that Ratch has a great way of explaining things correctly, with the right answer. I don't doubt he could work the problem without references though, and I should have thought before posting the second time, my first answer seemed to easy.
     
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