# Pussy Beware!

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

1. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
5,005
523
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
6,357
726
Thought this out, don't ask me for an equation.

108 (assuming no partial cats were slain)

3. ### wr8y Active Member

Sep 16, 2008
232
1
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 Expert

Jan 18, 2008
6,278
1,165
I got 972 cats.

John

5. ### Ratch New Member

Mar 20, 2007
1,068
4
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
6,357
726
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. 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
5,005
523
You folks were quicker off the mark with this one.

Well done Ratch

Edit: sorry and John.

Last edited: Mar 4, 2009

Mar 4, 2009
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9. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
5,005
523
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 AAC Fanatic!

Feb 19, 2009
6,357
726
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
2
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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 Expert

Jan 18, 2008
6,278
1,165
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
444
0
can someone call the PETA guys pleazzzz ?

14. ### wr8y Active Member

Sep 16, 2008
232
1
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
6,357
726
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.