Pussy Beware!

Thread Starter

studiot

Joined Nov 9, 2007
5,003
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.
 

thatoneguy

Joined Feb 19, 2009
6,349
Thought this out, don't ask me for an equation. :eek:

108 (assuming no partial cats were slain)
 

wr8y

Joined Sep 16, 2008
232
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?
 

Ratch

Joined Mar 20, 2007
1,068
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
 

thatoneguy

Joined Feb 19, 2009
6,349
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
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.
 
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Thread Starter

studiot

Joined Nov 9, 2007
5,003
You folks were quicker off the mark with this one.

Well done Ratch

Edit: sorry and John.
 
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Thread Starter

studiot

Joined Nov 9, 2007
5,003
Welcome to AAC, FX12

Old but good question.
I didn't claim originality and I didn't find it on the internet. The ditty is much much older than the net.
 

fx12

Joined Mar 4, 2009
2
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?
 

jpanhalt

Joined Jan 18, 2008
7,683
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
 

wr8y

Joined Sep 16, 2008
232
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
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....
 

thatoneguy

Joined Feb 19, 2009
6,349
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?
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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