if you were to google my exact question you'd find the answer fairly fast

 Ahh, of course.

I doubt people here do that though, otherwise they would've found the answer to your impossible one (I'm on my 4th or 5th hour trying to figure it out now) fairly fast, wouldn't they?

No, not really. I changed that one up a bit, and it took me a little while to find it because the person who gave me the problem changed it up a little bit too. The only advantage I have, is that the TA for lab told me the name of the problem after I figured it out and I managed to remember enough of it.

what is the super tuff one, i dont have time to scroll up for it

This one mesoteto.

i think outside of teh box and since the rules didnt say i couldnt

there i found a way to get there

I don't think that counts

ok, a new problem for those who read this thread

You have a pile of 24 coins. Twenty-three of these coins have the same weight, and one is heavier. Your task is to determine which coin is heavier and do so in the minimum number of weighings. You are given a beam balance (scale), which will compare the weight of any two sets of coins out of the total set of 24 coins. How many weighings are required to identify the heavier coin?

12 on each side for the first weighing.  12 are eliminated.

6 on each side for the second weighing.  6 are eliminated.

3 on each side for the third weighing.  3 are eliminated.

1 on each side for the fourth weighing. 

If one is heavier, that's the one.  If they are the same, the one you didn't weigh is the heavy one.

 Hmm. I'm saying 3.

First I'd break them into 3 piles. If it goes even, neither has the heavier coin. If it goes uneven, you know where it is. Either way, you know which 8 pile you have the heavier one in.

Secondly, you do the same with 3-3-2. Weigh 3-3, and have the same concept as in 1.

Third. If it was in the 2 group, simply weigh them against each other. If it was in the 3 one, weigh 2 against each other, if it balances, it's the last one.

Hey, mine was better than WoW's!_?