riddles...

[GoldPlatedDish]

My signature used to contain the following, I'll put it here as an enigma or something to ponder about:

Happiness comes in groups of four, and leaves in groups of ten.
Revel in your mistakes and failure shall ensue.
For maximum happiness, try to persuade a group of 40 to leave
With the tall one; he is quite the negotiator.

What am I talking about? THE DEVIL LORD OF ALL THINGS AWESOME AND COOL WANTS YOU TO TELL THE ANSWER, MORTALS!!!
*ahem* *sorry* *ahem*

[Ando]

My signature used to contain the following, I'll put it here as an enigma or something to ponder about:

Happiness comes in groups of four, and leaves in groups of ten.
Revel in your mistakes and failure shall ensue.
For maximum happiness, try to persuade a group of 40 to leave
With the tall one; he is quite the negotiator.

What am I talking about? THE DEVIL LORD OF ALL THINGS AWESOME AND COOL WANTS YOU TO TELL THE ANSWER, MORTALS!!!
*ahem* *sorry* *ahem*

are you serious? it's so obvious, its pokeman!!

[GoldPlatedDish]

are you serious? it's so obvious, its pokeman!!

FAIL!
But nice try, here's a cookie

[jujube]

[Paul676]

I C be eh to work it out.

[zaphod77]

GPD is obviously talking about Tetris Sprint.

If a = 0 and b=0, then c=0 solves both equations.

If we assume there is a unique solution, then 0 must be it. my answer is 0.

If not, then there is no answer.

I suspect it is solvable by graphing in 3 dimensional space and checking for the intersecting points, of which 0,0,0 is definitely one of.

Gah. a lot more algebra, and the solution a=b=c=0 is confirmed.

Will show work on request.

[solo2001]

Alright, I have no idea 'bout the GPD riddle, but how about this!?
[div align=\\\"center\\\"]What happens when Pinocchio says[/div]
[div align=\\\"center\\\"]"My nose will grow"?[/div]

[zaphod77]

Nothing happens.

He didn't say it would grow NOW.  If he is planning to lie in the future, than the statement is true, and his nose stays small.

It growing would be paradoxical at this point, though.

[jujube]

GPD is obviously talking about Tetris Sprint.

If a = 0 and b=0, then c=0 solves both equations.

If we assume there is a unique solution, then 0 must be it. my answer is 0.

haha can't believe i didn't think of that. i have to give you credit. that is definitely a valid answer.

[!--quoteo--][div class=\\\'quotetop\\\']QUOTE[/div][div class=\\\'quotemain\\\'][!--quotec--]
If not, then there is no answer.
[/quote]
oh but there is..and you don't have to plug numbers if you don't want to. it's coded into the riddle in a few ways.

i'm really glad you pointed out the 0 answer though. the next time i show this riddle i'll have to put in another line somewhere.

actually there is something else you should know, and i neglected to show it before. this actually does negate the 0s answer:
a÷b=c

edit: there are still two answers for C. the absolute values for A, B, and C are shuffled between the variables when comparing both answers, but the same three absolute values are present in each valid answer (does that make sense?).

and zaphod i see what you mean by "If not, then there is no answer", meaning that if the answer wasn't 0 then the riddle becomes untrue, as it says there's only one answer. but i see now you didn't necessarily mean there couldn't be other answers.

anyway sorry about the technicalities. one day i'll get the wording of everything right, so that there can be no zero or negative answers. i still like the riddle a lot in this exact form, so i'm not sure how i would change it.

[EnFuego]

GPD is obviously talking about Tetris Sprint.

If a = 0 and b=0, then c=0 solves both equations.

If we assume there is a unique solution, then 0 must be it. my answer is 0.

If not, then there is no answer.

I suspect it is solvable by graphing in 3 dimensional space and checking for the intersecting points, of which 0,0,0 is definitely one of.

Gah. a lot more algebra, and the solution a=b=c=0 is confirmed.

Will show work on request.

C can be 0 but I think the answer was what must C be. If a and b are additive inverses (b=-a) then then c=0 is invalid.

[xlro]

so you say, we have:

(I) a + b = c
(II) a / b = c
(III) ab + bc = ca

from (I) + (II) we can see that b and therefore c and a can't be 0 (neither one)

using (II) [a = bc] in (III) gives c(b+1) - c^2 = 0, so this only holds for c = b + 1, this means a = 1.

so in (III): c = b / (1 - b ), or c = (c-1)/(2-c) => c^2 - c - 1 = 0
this has the two solutions c = 0.5 +/- 0.5*sqrt(5)

and that's what c must be, eh.
as stated above, you get b = c - 1

btw, if you leave out (II) the solution is ofc not totally defined, you can just give it in dependency of a.

[jujube]

that is excellent xlro  and you're right about the consistent value of a. however, "1" is a boring number!
0.5 + 0.5*sqrt(5) is the answer i was looking for, and i especially like this way of expressing it. i'm sure a lot of you know what this number is.

[xlro]

at first I didnt see it, but it's really a nice solution
one could directly see it, if you bring it in the form:
a / c = c / (a + c)
for a = 1, c is the golden ratio

[zaphod77]

Are you telling me you left out an equation? because there were only two in the original problem.

A+B=C
AB+BC=CA

and now you are adding on

A/B=C

That definitely affects things.

And yeah, that gives the answer you expected.  Not my fault you left out an equation.

[xlro]

a/b = c is not really needed, the golden ratio is already given bei the two initial equations, since

a + b = c
ab + bc = ca

still has the solutions c = a * (0.5 +/- 0.5*sqrt(5))
(besides the trivial a=b=c=0 one you mentioned)

so it's just scaled by a (= 1), but describes the same thing