Puzzle Time!

Dang! I thought you meant all the basic order of operations. Third time's the charm, eh?

Well I hope this one finally works. [spoiler](888/.888)/(8/8)[/spoiler]

Haha, that's a cheat. But a nice one!

New one! You have 81 marbles. All are identical except one, which weights more than the rest. You have also have a [url=http://upload.wikimedia.org/wikipedia/commons/f/fa/Balanced_scale_of_Justice.svg]balance scale[/url]. What is the minimum number of measurements needed to determine the heavier marble?

Four

If you are lucky, one. If you need an algorithm that works in every situation, yes, four :) [spoiler] 27 27 27 9 9 9 3 3 3 1 1 1 [/spoiler]

Too easy for you guys ;)

A king needed to make gold coins for his kingdom, so he hired 10 goldsmiths. Each one was instructed to make each coin using .1oz of gold. However, he heard rumors that one of the goldsmiths was cheating him, using less than .1oz of gold in his coins! The king also has a scale which will give a numerical weight (like a bathroom scale). Using only one measurement, how can the king determine which of the goldsmiths is cheating him?

Nice one. [spoiler=Solution]He could measure 1 coin from goldsmith #1, 10 coins from goldsmith #2, 100 from #3, etc... If he sees the weight like this: 1110932146, than he knows that goldsmith #7 is cheating, as he is the first one from the left to have a weight less than 1 oz per coin. P.S. Don't use my bathroom scale, as it doesn't measure in oz :) [/spoiler]

Dominating12 has started to mint tokens! You are given x number of tokens (x>10) on a table and told that only 10 are heads up. With your eyes closed, how can you divide the tokens into two groups so that each group contains the same number of heads-up coins? (If you win, you get to keep the extra tokens). Hint: [spoiler]You may flip over any number of tokens[/spoiler] Hint II: [spoiler]Each group does not necessarily need to have 5 heads-up tokens; they only need to have an equal number of heads.[/spoiler]

[spoiler=answer] 1. Count all coins, lets say there are 13 coins, to make it easy to understand (it works equally wel if you say x=10+y, where y is the number of tails - in my case that's 3). 2. Put 3 coins in one group, and 10 coins in the other group, and flip all of them. 3. Now fil all coins in the left group again, and you are done. How it works: After step 2, we have three heads in total, these are the possibilities (in a table for group L(eft) and group R(ight)): L|R --- 3|0 2|1 1|2 0|3 - So in the first case, if we flip all 3 coins, there are only tails, so 0 heads in both groups. - In the second case, there is one head in the right group. In the left group, there are two heads and 1 tail. Flip all of those, than there is 1 head in the left group, and 1 in the right group. - ... :) [/spoiler]

[spoiler]Nice Matty :) Wouldn't it be easier, though, to simply flip all the coins in the group with 10?[/spoiler]

Yes, that would be easier, but this way it's easier to understand why I'm doing this.

Ok simple poker quiz two-part question 1) Can you make a straight without a 10 or a 5?, 2) Are there any straights you can't you make in Texas holdem with 10 5 as your starting hand?

1) Not unless you are playing with wild cards. In a standard game (no wild) you can not make a straight without either a 5 or a 10. Unless you are playing some wacky game that allows a "wrap around" straight (QKA23). But those are silly kids games and not common. 2) No, all straights are possible if you are dealt a 5 & 10. In fact all straights are possible (in theory) no matter what you are dealt. If you have a 2 & 3 in the pocket, is it possible to make a 7 through Jack straight? Yes, if the 7, 8, 9, 10, J are all turned over as the shared common cards... So no matter what you are dealt, all straights are possible.