Send kids.
Send one kid back.
Send fatso.
Send other kid back.
Sends kids.
I guess that one everyone knows 
How about sending a lion, a goat and straw?
You cannot leave the goat with the straw or the goat will eat it, and, for the same reason, you cannot leave the lion with the goat.
BTW, you can only carry one thing at a time.
Send goat.
Go back.
Send lion.
Bring goat back.
Go back.
Send straw.
Go back.
Send goat.
Strangelove just burns through these riddles with his avatar flamethrower.
actually hes aiming a bit higher so he missed, doesn’t surprise me since hes a noob
You are driving a bus. At your first stop, you pick up 8 people. The next stop, 4 people get off and 2 people get on. The next stop, 3 people get on and no one gets off. The next stop, 9 people get on and 8 people get off. The next stop, no one gets on and one person gets off. The next stop, 4 people get on and no one gets off. The next stop, 9 people get on and no one gets off. The next stop, 8 people get off and no one gets on. What is the bus driver’s name?
^ That one is funny because everyone does the math and forgets the beginning. It isn’t that fun when you write it down, though.
Anyway, this one is actually tricky:
A bear walks 30 km to the South, 30 km to the East and another 30 km to the North. It ends up exactly where it began to walk.
What is the color of the bear?
White?
Oh, dammit, I’ll have to search some harder problems :[
Yes, you will.
(I, being the pro that I am, answered the questions as I scrolled down)
1: x+x+x+x…+x = x^2
2: (d/dx)(x+x+…+x) = (d/dx)(x^2)
3: (d/dx)x+(d/dx)x+…+(d/dx)x = 2x
4: 1+1+…+1 = 2x
5: x = 2x
which line is wrong?
/* that is required the calculus experiences for solving the problem. */
1: x^4 + x^3 + x^2 + x + 1 = 0 // define
2: x(x^3 + x^2 + x + 1) + 1 = 0
3: x^3 +x^2 + x + 1 = -x^4 // from 1:, minus x^4 from each side
4: x(-x^4) + 1 = 0 // from 2: and 3:, insert whole polynomial by -x^4
5: x^5 = 1
6: x = 1 ------------> from 1: : 1+1+1+1+1 =0 ???
which line is wrong?
/* that is required the algebraic experiences for solving the problem. */
If i’m not mistaken the path describes the North pole, not the South pole, but there aren’t polar bears North.
In the first example the first line is only valid only for x being a natural number and a function defined on the natural number set isn’t differentiable.
In the second example you are mixing equations with functions - the equation in the 3. line have no solutions, so the function on the left side f(x) = x^3+x^2+x+1 isn’t equal to the function on the right side g(x)= -x^4, so one cannot be replaced with another.
There aren’t? Damn you global warming! :’(
then clearly the bear is a penguin.
This solution works much better for the original version of this puzzle…
Both problems have the same answer. To prove that the solution satisfies my problem as well, one should also prove that the solution is the “fastest solution”, but that can be done by induction.
The original version of the problem didn’t help me think, though, because people won’t do some reasoning together if they don’t agree on it.
Plus, I’m not sure but I think the resolution is actually slightly flawed. I can’t explain why without sound confusing, but the only way to convince me would be to formalize the problem and prove it by induction.
Anyway, if people could agree on how to behave “before the games starts”, then the formalization is simple using Turing machines, with the agreement being the algorithm they follow.
You are mistaken.
I don’t want to sound rude, but I don’t think your answer is right.
Line [3] is equivalent to line [1], so saying line [3] has no solution is equivalent to saying the problem itself has no solution.
I would rather say that the line of reasoning is right. You begin the problem assuming there is a real number x such that the equation in line [1] is valid. If there is such an x then, following the steps, x must be 1. But the number 1 does not satisfy the first equation, therefore what you conclude is that equation [1] has no real solution.
In other words, he proved that x^4 + x^3 + x^2 + x + 1 = 0, with x being a real number, implies x = 1. This is right. The reverse implication is not right, but that doesn’t matter, that is not what was proved and the steps are not reversible.
On a side note, if you allow x to be a complex number, then the step 4 -> 5 is wrong. What you would conclude is that x must be one of the five complex solutions of x^5 = 1. One of them is 1 and the other four are the solutions of equation [1].
Yeah, but there was a few more tips in the original that helped solving the puzzle - for instance all participants were logicians, “the bests of the bests”, so you could assume that all of them would follow logical rules. Additionally, there were also other colors and you were told what was going on on each iteration which made the solution a little more obvious (well, actually it was still hard as hell to figure it out).
Hey, there’s nothing rude in pointing errors, especially when you are a student of Maths and I’m not 
Though, correct me if I’m wrong, but [3] is true <=> x є ø, so [4] should look like this:
- x(-x^4) + 1 = 0 AND x є ø
so when you have the final solution you would get:
x = 1 AND x є ø <=> x є ø
A bit late there, babe.
First; you are wrong. x+x+x…+x (for x times) is differentiable but you have to change that from into algebraic form : x^2.
to prove this:
(d/dx)(x) = 1
(d/dx)(2x) = 2
(d/dx)(3x) = 3
(d/dx)(xx) ≠ x, get it?
Second; as like as what gugamilare said. you can put function(is it?) into expression -x^4, that still right. but, the point is this equation have no answer in REAL line—> x^5 -1 = 0 doesn’t mean that x=1.
x^5-1 have five factor : (x-1)(x-Z1)(x-Z2)(x-Z3)(x-Z4)=0
for the complex Z1,Z2…,Z4 that can write in this from : a+bi , a,b є R and i=sqrt(-1).
and we’ve seen that 1 doesn’t the solution so another Z1,Z2…Z4 does.
you got it again 