@Johann : You were right this brainteaser is really hard. I don't know where to begin to solve it.

If you don't mind, I'll add another brainteaser to keep the thread going while we all try to figure it out yours (and nordlead's puzzle too!)

So here's mine :

You have 25 horses who all run at a constant speed but not one of them at the same speed. The track only has 5 lanes so you can only race 5 horses at a time. What's the least amount of races that you need to identify the 3 fastest horses.

I`m glad I`m not the only one who thinks so! I lost so many nights of sleep over the damn thing, I was starting to think that maybe I was just too dumb to understand it!

I`ll give your horse problem a shot:

5 races: Race all the horses (using a timer) and compare the lap times of the winners to find out who the fastest are.

That's cheating! :) You don't have a timer so all the information you can get from the races is the order they come in.

@ TruckOSaurus: That`s really the sort of thing you have to specify on the description of the problem! It`s not cheating if you didn`t warn me.

But the fact that your got so worked up over that is a clue...
If the least ammount of races was a number smaller than 5, I guess you would simply tell me that I am wrong instead of acusing me of cheating. So the number of races has to be 5, because 6 is the most obvious (wrong) answer.

So based on those facts, my guess is 5 is the lleast ammount of races! Am I right?

Johann said: @ TruckOSaurus: That`s really the sort of thing you have to specify on the description of the problem! It`s not cheating if you didn`t warn me. But the fact that your got so worked up over that is a clue... If the least ammount of races was a number smaller than 5, I guess you would simply tell me that I am wrong instead of acusing me of cheating. So the number of races has to be 5, because 6 is the most obvious (wrong) answer. So based on those facts, my guess is 5 is the lleast ammount of races! Am I right?

I'm gonna assume that you meant you separated all 25 horses into groups of 5 and that you raced them all in your 5 races but that wouldn't be enough for you to figure out the three best horses because each horse is being compared to only 4 other horses.

Ok, no one is posting here! This thread can't die!

We have to keep it going!

Nordlead's problem is too much for a human being to handle!

As for TruckOSaurus's horses one... I'm gonna guess 6 races just to get it out of the way: divide the horses in groups of five and then race the 5 winners

My guess it that isn't enough to guess the 3 fastest, say you group them by chance them this way:

1st race: 1st quickest-2nd quickest-3rd quickest-4-5

2nd race: 6-7-8-9-10

and so on, you'd end with 1-6-11-16-21 racing the final, and you'd end with 1-6-11 as the quickest horses

The safest bet is to race any horses, pick the three fastest, race them against two more, pick the three fastest and so on. That way you can be sure that you'll end with the three fastest, and it will take you 11 races. I don't know if there's one quicker way to do it, though

Nordlead, short answer: Not possible.

You're working with multiples of 2 and trying to divide them by 3. It's simply put not possible. I'm sure you would like an explanation or a proof. I just cannot give you that :( But for the sake of my argument I will try.

I'm am uncertain about whether the position of the B's and C's actually matter, but I'll try to move on as if they don't (because they really don't. Really. And even if they did matter, if it's impossible with the looser set of rules, it's even more impossible with the more strict set).

We are given a couple of tools. First tool is: multiply by 2 (Vx >> Vxx).
Second tool is Divide by three (exchange 3 G's with a C).  Actually, this is not a tool at all. It's only an imaginary tool that deals with the illusion that G's and C's are seperate entities and not numbers.
Third tool is add 3 (you can add C to the end of the string)
Fourth tool is remove 6 (two C's in a row can be removed).
You start with 1G and the goal is to end up with 3G, since 3G is equal to C.

The only way for this to happen is if you could somehow end up with a multiple of 3. Therefore I will look at and see if any of these tools could make this happen.

If you start with a number that is not a multiple of 3, and you can multiply it with 2, you can easily show mathematically that you will never end up with a multiple of 3. If you add 3 to a number not divisible by three you will never end up with a number divisible by three. If you subtract 6 from a number not divisible by three you will never end up with a multiple of 3.

Since you start with a number not divisible by 3, and none of the rules lets you get from a number not divisible by 3 to a numver that is divisible by three, it is impossible with the current rules to go from 1 to 3.

... and it's 7 in the morning now. I hope I have arrived at the right conclusion, but some evil detail could have eluded me...
If so I won't stand by my mistake :)

EDIT: I edited my answer a bit, to straighten out some answers and to make my reasoning (I hope) a bit clearer.

EDIT2: Eh. Where did I get the idea of B's from? It was G's. Argh!

Is this true??? Is it really not possible??? :(......

I'm not 100% sure it's not possible. I just cannot see how it could be. Try following my reasoning for it, and see if you can find a flaw or three in there.