A wealthy man owned a stable of 'seventeen' horses. When he died he bequeathed the horses to his three sons. The will stated that eldest son was to be given one half of the horses, the middle son was to be given one third of the horses, and the youngest son was to be given one ninth of the horses. The sons were distraught. It was clear to all that the horses could not divided in this way without making a bloody mess.

Question: How can they divide the horses as per old man's will? Give it a thought before looking at the answer! (http://www.gigglepedia.com/puzzle/horse-paradox.asp). You will be just amazed with the solution.

Add one horse, and at the end you'll have one left. ;)

Shouldn't have included the link so I could've been a legitimate smartass here. :D

The only problem is, that the solution is incorrect. In the end, the oldest son will have 53% of the horses (instead of 50%), the middle son has 35,3% (instead of 33,3%) and the youngest has 11,7% of the horses (instead of 11,1%)
The father must have died of alzheimer anyway, because 1/2+1/3+1/9 is 8/9 and not 1.

The puzzle is technicly not logicly possible to answer because 1/2, 1/3, and 1/9 are not perfect divisors of 17.

It's mathematically impossible to solve this riddle like Gigaz said.

1/2 + 1/3 + 1/9 = 9/18 + 6/18 + 2/18 = 17/18.

That's why you add the horse to make it eighteen, divvy up the horses, and take back the one you added.

fwiw, I have this same puzzle in a book of ancient Sufi stories, so it's been around a long time.

Adding another horse to the mix does not solve the problem. The problem asks how to divide seventeen horses, not eighteen horses.

Sell them all and divide the money.

Adding another horse to the mix does not solve the problem. The problem asks how to divide seventeen horses, not eighteen horses.
You add one then take it away still having 17 so you don't need the extra horse.

Sell them all and divide the money.

And then use the money to buy new horses.

I find several of the responses funnier than the paradox.

You add one then take it away still having 17 so you don't need the extra horse.

No that's retarded, he was rite. If we go by those rules then I say, chop up the horses and bring in some scientists from the future to animate the limp pieces, see, by this stupid ass answers logic I could totally do that.

Our math teacher told us this riddle in middle school, I solved it then. Although I didn't think of adding one. I simply knew that the answers were the fractions of 18 instead of 17, and ergo a larger percentage of 17.

No one laughed, btw. Half the class knew the answer by the time I decided to speak up.

You add one then take it away still having 17 so you don't need the extra horse.

That doesn't matter. The father's will specified that his eldest son get half(50%) of his horses. Instead, he got around 53% of his horses. That's going against the father's will, and thus, wrong.

I'd breed them until a suitable number, then divide.

Shoot 'em all and let God sort it out.

Shouldn't the sons be distraught about their father dying rather than the will being impossible to fulfill without the help of a glue factory?

Shouldn't the sons be distraught about their father dying rather than the will being impossible to fulfill without the help of a glue factory?

No, he had suffered long enough, and it was a blessing to finally see him go.

Besides, he was always torturing his sons with impossible requests like how to split up these darn horses, and it was only getting worse as he got older. Some of these requests seemed completely ridiculous, so they were kind of relieved when he died. The last straw was that his will also stated that he should be buried face-down so the whole world can bend over and kiss his a**.

Yeah, that's a stupid "paradox" that could have been a clever riddle if it was worded better.

I think there are 3 possibilities:
1) One of the horees is preganent.
2) THe guy was going to get more horses.
3) The father wanted his sons to have a massive feast.

This doesn't make any sense at all. I want to commit suicide!

Shoot 'em all and let God sort it out.
:rotfl: :rotfl:

Another good one which proves that x=0:

x=y

x2=xy Multiply by y

x2 - y2 = xy - y2 Subtract y2

(x-y)(x+y) = y(x-y) As x2 - y2 = (x-y)(x+y

(x+y)=y Cancel the (x-y)s

x=0 Remove the brackets and simplify

you're dividing by 0 when your canceling the (x-y)

That didn't take long... OK, another one:

3>2

3 log(1/2) > 2 log(1/2)

log (1/2)^3 > log (1/2)^2

(1/2)^3 > (1/2)^2

1/8>1/4

thats not a parradox, just a series of incorrect statements. You can only keep an equality after having done an opperation to both sides of it, if that opperation is a continuously increasing functio.
ln(1/2) is negative, so the second line is wrong, and the one's below that are too.

If we're doing fake proofs, my favorite is:
4-10=9-15
4-10+25/4=9-15+25/4 --add 25/4 to both sides
(2-5/2)(2-5/2)=(3-5/2)(3-5/2) --factor both sides
2-5/2=3-5/2 --square root both sides
2=3 --add 5/2 to both sides

thats not a parradox, just a series of incorrect statements. You can only keep an equality after having done an opperation to both sides of it, if that opperation is a continuously increasing functio.
ln(1/2) is negative, so the second line is wrong, and the one's below that are too.

Correct: that's the big flaw. By the way, there is no method of getting from a logical premise via a series of logical steps to an illogical conclusion, so every paradox is just a series of incorrect statements

If we're doing fake proofs, my favorite is:

Code:
4-10=9-15
4-10+25/4=9-15+25/4 --add 25/4 to both sides
(2-5/2)(2-5/2)=(3-5/2)(3-5/2) --factor both sides
2-5/2=3-5/2 --square root both sides
2=3 --add 5/2 to both sides



Square Root both sides is an invalid operation as you are actually multiplying/dividing (depending on how you look at it) by different numbers on either side

Square Root both sides is an invalid operation as you are actually multiplying/dividing (depending on how you look at it) by different numbers on either sideThat's not quite it. There is no multiplying or dividing, and even if there were, it would be by equivalent expressions. Taking the square of both sides would certainly be valid, because you are just multiplying both sides by equivalent expressions.

You are right that that is the trouble line, and that taking the square root is not done this way, but your reasons for why not is wrong.

What set of two numbers can you multiply to get (3-5/2)(3-5/2)?

If we're doing fake proofs, my favorite is:
4-10=9-15
4-10+25/4=9-15+25/4 --add 25/4 to both sides
(2-5/2)(2-5/2)=(3-5/2)(3-5/2) --factor both sides
2-5/2=3-5/2 --square root both sides
2=3 --add 5/2 to both sides

Square root of x = x^1/2 which is a multiplication or a division depending on how you look at it. Supposing it was x and y ^3, and x was 2 and y 3, then it would be 2*2*2 = 8 against 3*3*3 = 27

Square root of x = x^1/2 which is a multiplication or a division depending on how you look at it. Supposing it was x and y ^3, and x was 2 and y 3, then it would be 2*2*2 = 8 against 3*3*3 = 27
You need to start with an equality For example this is valid for all x, y:
x=y^3 -> x^2=(y^3)^2=y^6

Specifically try (x=1, y=1), (x=8 y=2), and (x=3, x=2).

Multiplying, dividing, adding, and subtracting can both be done by applying the operation to both sides by with representations of equal expressions. This is essentially the same as multiplying by the same expression, then substituting it on one side for the equivalent alternative.

But taking the square root is not multiplication or division.

x^1/2 means "x multiplied by itself half a time", does it not? Anyway, it seems that one has been done to death a bit - any more?

That's just the thing, it doesn't. If it did, then it would be valid, just like taking the square is.

The problem is that there are two answers to taking the square root, the positive, and the negative root. In our problem 3-5/2 is the positive root (1/2), and 2-5/2 is the negative root. (-1/2) Therefore if you want to take the square root of both sides, you have to also take the absolute value of the result. This absolute value would render the last step invalid.

When you're square rooting, you can take either the positive or the negative square root. In that paradox your taking the negative root on one side, and the positive on the other side. Took me a while to figure out mind you.

That's just the thing, it doesn't. If it did, then it would be valid, just like taking the square is.

The problem is that there are two answers to taking the square root, the positive, and the negative root. In our problem 3-5/2 is the positive root (1/2), and 2-5/2 is the negative root. (-1/2) Therefore if you want to take the square root of both sides, you have to also take the absolute value of the result. This absolute value would render the last step invalid.

Oh, OK. I only did O-level maths...