mental challenge

[bunsen burner]

Test your mental agility with the following puzzle:

Paul and Ross have been cat-breeders of renown for many years. Their insight into the nuances of felines have won them much acclaim. However, they've always been allergic to cats and have finally decided that it is foolishness of the highest order to remain in the cat-breeding business.

They decide instead to venture boldly into dog-breeding. So they take all their cats to the local cat markets, and sell them off. In a coincidence surely worthy of any logic puzzle, the amount in dollars they receive for each cat is exactly the same as the number of cats they sold.

For this money they purchase as many dogs as possible at $10 a dog. And because they're a little bit zany, they spend the left-over money on a flea circus.

On their way home they have a massive fight about the merits or otherwise of swing music and decide to split up forever. Unfortunately, when they divide the dogs there is one dog left over. So Paul says to Ross "I'll claim the last dog and you can have the flea circus." "Hold your horses, Swing Boy," says Ross, "The flea circus costs less than a dog". "Ok", Paul says. "Then I'll give you one of my skateboards and we'll be even". And Ross agrees.

How much was the skateboard worth?

 

[otaku]

skateboard = (1dog-1flea circus)

 

[evade28]

well since the cats sold for one dollar each and the flea circus cost less then a dog, which were $10 each then the skateboard is worth anything between 1 and 9 dollars?

 

[the fly]

The skateboard was worth $4

 

[evade28]

Originally posted by the fly
The skateboard was worth $4

how did u come up with that figure?

i forgot to count the flea circus in my figures so i would say the skateboard would be worth anything between one and eight dollars

 

[El Scorcho]

Yeah, I get $4 too.

Originally posted by evade28
how did u come up with that figure?

evade28, read the following bit carefully;

Originally posted by Bunsen Burner
In a coincidence surely worthy of any logic puzzle, the amount in dollars they receive for each cat is exactly the same as the number of cats they sold.

 

[evade28]

Originally posted by El Scorcho
Yeah, I get $4 too.



evade28, read the following bit carefully;

hmmm... i thought it was like one dollar for one cat, so $10 for 10 cats?

but now that u mention it i see that if they sold 7 cats they would have sold them for $7 each, riiite hehe!

now my theory is farkd!

 

[Darky]

I've tried the puzzle with four different numbers of cats, and the end result is always the skateboard being worth $4.

 

[evade28]

must be a pretty sh1t skateboard then!!

can someone explain how they got 4 bucks?

 

[Darky]

Originally posted by evade28
must be a pretty sh1t skateboard then!!

can someone explain how they got 4 bucks?

I'll PM you the explanation, incase others are still trying to work it out on here.

 

[Korn]

yes i got $4 as well
got any more? i love doing these problems

 

[bunsen burner]

I'm pretty sure the answer isn't $4.

Darky, can you post your explanation?

I got this sent to me with Residex.com's latest newsletter. They don't give the answer until next month. I got something different to $4.

 

[Qidds]

Originally posted by bunsen burner
I'm pretty sure the answer isn't $4.

Darky, can you post your explanation?

I got this sent to me with Residex.com's latest newsletter. They don't give the answer until next month. I got something different to $4.

I get $2 for the skateboard. I think you have to remember that the skateboard isn't actually shared to begin with, so skateboard = 1/2 (dog - flea circus).

It's interesting that the answer (mine, at least) is independent of the number of cats. When the square of the number of cats is divided by 10, you have to be left with an odd number plus a remainder - by some beautiful arithmetic quirk, the remainder is always 6 (6^2 = 36; 14^2 = 196; 16^2 = 256). All other squares leave an even number when divided by 10 (until the square becomes 4 digits and my head begins to hurt....).

 

[Korn]

Originally posted by Qidds
I get $2 for the skateboard. I think you have to remember that the skateboard isn't actually shared to begin with, so skateboard = 1/2 (dog - flea circus).

It's interesting that the answer (mine, at least) is independent of the number of cats. When the square of the number of cats is divided by 10, you have to be left with an odd number plus a remainder - by some beautiful arithmetic quirk, the remainder is always 6 (6^2 = 36; 14^2 = 196; 16^2 = 256). All other squares leave an even number when divided by 10 (until the square becomes 4 digits and my head begins to hurt....).

i dont get that 1/2 thing for the skateboard.
the way i worked out is the same as you mentioned though.

 

[Darky]

Same theory as the two posts above.

This is the PM I posted to evade28 :

------------------------------------------------------------------------------------
Okay, I did this using trial and error to find the right number.

For the number of cats, pick a number, then multiply it by that number.

So (eg) 8 cats x $8 = $64

This would buy 6 dogs @ $10 each, with $4 remaining.

The puzzle says that when the two guys split up, there is one dog remaining... so there couldn't have been 6 dogs because it's an even number. So 8 is not the right number of cats.

Try it with a few different numbers :
4 x 4 = 16. 1 dog, $6 remaining
5 x 5 = 25. 2 dogs (even) puzzle doesn't work
6 x 6 = 36. 3 dogs, $6 remaining
7 x 7 = 49. 4 dogs (even) puzzle won't work
8 x 8 = 64. 6 dogs (even)
9 x 9 = 81. 8 dogs (even)
10 x 10 = 100. 10 dogs (even)
11 x 11 = 121. 12 dogs (even)
12 x 12 = 144. 14 dogs (even)
13 x 13 = 169. 16 dogs (even)
14 x 14 = 196. 19 dogs, $6 remaining
15 x 15 = 225. 22 dogs (even)
16 x 16 = 256. 25 dogs, $ 6 remaining
17 x 17 = 289. 28 dogs (even)
18 x 18 = 324. 32 dogs (even)
...and so on

So with an even amount of dogs, the clues for the puzzle don't match.

Wherever there is an odd number of dogs, there is always $6 remaining.

If the flea circus + the skateboard = same value as the odd-numbered (last) dog, and the flea circus is $6 and the dog $10... then the skateboard must be worth $4.

Yes, that would make it a cheap, **** skateboard.

------------------------------------------------------------------------------------

 

[Qidds]

Originally posted by Darky
Same theory as the two posts above.

This is the PM I posted to evade28 :

------------------------------------------------------------------------------------
Okay, I did this using trial and error to find the right number.

For the number of cats, pick a number, then multiply it by that number.

So (eg) 8 cats x $8 = $64

This would buy 6 dogs @ $10 each, with $4 remaining.

The puzzle says that when the two guys split up, there is one dog remaining... so there couldn't have been 6 dogs because it's an even number. So 8 is not the right number of cats.

Try it with a few different numbers :
4 x 4 = 16. 1 dog, $6 remaining
5 x 5 = 25. 2 dogs (even) puzzle doesn't work
6 x 6 = 36. 3 dogs, $6 remaining
7 x 7 = 49. 4 dogs (even) puzzle won't work
8 x 8 = 64. 6 dogs (even)
9 x 9 = 81. 8 dogs (even)
10 x 10 = 100. 10 dogs (even)
11 x 11 = 121. 12 dogs (even)
12 x 12 = 144. 14 dogs (even)
13 x 13 = 169. 16 dogs (even)
14 x 14 = 196. 19 dogs, $6 remaining
15 x 15 = 225. 22 dogs (even)
16 x 16 = 256. 25 dogs, $ 6 remaining
17 x 17 = 289. 28 dogs (even)
18 x 18 = 324. 32 dogs (even)
...and so on

So with an even amount of dogs, the clues for the puzzle don't match.

Wherever there is an odd number of dogs, there is always $6 remaining.

If the flea circus + the skateboard = same value as the odd-numbered (last) dog, and the flea circus is $6 and the dog $10... then the skateboard must be worth $4.

Yes, that would make it a cheap, **** skateboard.

------------------------------------------------------------------------------------

But the skateboard originally belongs to Paul - it's not shared (like the dogs and flea circus). The dogs and the flea circus are originally half-owned by both Ross and Paul, the skateboard isn't.

Say your grandmother gave you and your brother $10 to share, how much would he have to pay you to let him have it all ? $5, not $10.

So, the dogs are worth $10 and the flea-circus $6. Paul takes a dog from the share .

Paul up $5 ; Ross down $5

Ross takes a flea circus from the share

Paul up 5-3 = $2; Ross down 5-3 = $2.

Now Paul must give something (not from the share) to Ross to make up for it. Thus the skateboard is worth $2.

If the skateboard were worth $4, Paul has gotten half a dog ($5), but lost half a flea circus ($3) and $4 worth of skateboard -> the deal wouldn't be even.

 

[Darky]

Originally posted by Qidds
But the skateboard originally belongs to Paul - it's not shared (like the dogs and flea circus). The dogs and the flea circus are originally half-owned by both Ross and Paul, the skateboard isn't.

Say your grandmother gave you and your brother $10 to share, how much would he have to pay you to let him have it all ? $5, not $10.

So, the dogs are worth $10 and the flea-circus $6. Paul takes a dog from the share .

Paul up $5 ; Ross down $5

Ross takes a flea circus from the share

Paul up 5-3 = $2; Ross down 5-3 = $2.

Now Paul must give something (not from the share) to Ross to make up for it. Thus the skateboard is worth $2.

If the skateboard were worth $4, Paul has gotten half a dog ($5), but lost half a flea circus ($3) and $4 worth of skateboard -> the deal wouldn't be even.

I understand... and realised it as soon as I saw your earlier post.

I just didn't think of it last night when I wrote the PM, at about 1 in the morning and after several stubbies consumed..

 

[red+black]

Originally posted by Qidds
But the skateboard originally belongs to Paul - it's not shared (like the dogs and flea circus). The dogs and the flea circus are originally half-owned by both Ross and Paul, the skateboard isn't.

Say your grandmother gave you and your brother $10 to share, how much would he have to pay you to let him have it all ? $5, not $10.

So, the dogs are worth $10 and the flea-circus $6. Paul takes a dog from the share .

Paul up $5 ; Ross down $5

Ross takes a flea circus from the share

Paul up 5-3 = $2; Ross down 5-3 = $2.

Now Paul must give something (not from the share) to Ross to make up for it. Thus the skateboard is worth $2.

If the skateboard were worth $4, Paul has gotten half a dog ($5), but lost half a flea circus ($3) and $4 worth of skateboard -> the deal wouldn't be even.

in the words of Kevin Bloody Wilson "Hey Mick, you're *ucken good!"

 

[Murph]

Either way, if the dogs were only worth $10 when they bought them, you can bet you house that they wont be worth crap because they can't breed, are three legged, have rabies, flea infested and probably gay, just like their owners.

Seems to me Ross is used to being dudded, wonder which person he played in the relationship with Paul, fancy taking a flea circus and a $2 piece of $h1t skateboard.

 

[Nic]

**** off, I did enough of this crap at school!

(I'm still asking myself why I felt it necessary to do Methods and Specialist)

 

[carn_the_nmfc]

I got $4 as well

 

[BomberGal]

Originally posted by Nic
(I'm still asking myself why I felt it necessary to do Methods and Specialist)

:|

I can't even contemplate why I asked my maths teachers if I could do both Methods and Specialist, let alone actually doing it. Methods is enough thank you. BTW, cbf doing that puzzle.

 

[NakedDeadGuy]

Got this emailed to me :

"Australian first!! Freecall to determine simply and easily if you are smart or stoopid! As easy as picking up the phone and dialling 0004 921530!!"

"**Don't call unless you've really thought about it, and are sure you are prepared for the possibly crushing news that you indeed are a tool**"

Don't mean to brag, but it at least proved that I'm not "stoopid".

 

[LongBombFromOutside50]

I haven't read any of the other answers and am about to post my answer, sorry if it has already been posted:

Since "the amount in dollars they receive for each cat is exactly the same as the number of cats they sold:"

T.C (Cats) = N (Cats) x N (cats)

T.C (Dogs) = A.C (Dog) x N (Dogs) = 10 x N (Dogs)

N (Dogs) > 2 and must be odd since the dogs are divided between Paul and Ross unequally.

T.C (Cats) - T.C (Dogs) = C (Flea Circus)

6 Cats: 36 - (10 x 3) = 6
14 Cats: 196 - (10 x 19) = 6
16 Cats: 256 - (10 x 25) = 6
24 Cats: 576 - (10 x 57) = 6
26 Cats: 676 - (10 x 67) = 6

Above are the only possible combinations of the N (Cats) between 1 and 30 given what we already know about N (Dog)


You could say:

C (Skateboard) = A.C (Dog) - C (Flea circus), therefore:

C (Skateboard) = 10 - 6 = $4

BUT, when you look back through the problem...

Assuming there are 16 cats (using one of the examples from the set of possible eqns above), the T.V (Cats) = 16 x 16 = 256

Since there are 25 dogs, " Paul says to Ross 'I'll claim the last dog and you can have the flea circus,' " therefore Paul keeps 13 dogs and Ross keeps 12 dogs:

T.V (Paul) = N (Dogs) x A.C(Dog) = 13 x 10 = 130

T.V (Ross) = N (Dogs) x A.C(Dog) + C (Flea circus) = 12 x 10 + 6 = 126

This means that the exchange in value between the two must equal $4, so the cost of the skateboard therefore = $2

We are now left with:
Total value of goods to Paul: $128
Total value of goods to Ross: $128


Shorthand abbreviations:

T.C = Total costs

T.V = Total value

C = Cost

A.C = Average Cost

N = Number of

 

[the fly]

So $2 it is. I forgot to take the comparative share of the money used to buy the dogs into account.