May I suggest this riddle?

If you can stand another riddle, hereis one I have always liked!
You are condemmed to death, but are given a chance. You are given a very precise balance (the kind that has two balancing trays) and nine pills. Eight of the pills are poisoned, and only one is safe to take. All nine pills are exactly the same in colour, look, size and weight, except that the safe pill is a bit heavier than the rest. You are only allowed to use the balance twice, and then you will have to eat one pill. What do you do?

how can they all be the same size and weight, yet the safe one still be heavier than the rest?

For the purposes of this riddle, they all are... c'mon, play along! lol ;)

I say, dont eat one, the riddle never told u that u had to take one... or am i notplaying right>

I would weigh 4 pills on each side for the first try. If there was any differences I would eliminate the 4 ones on the lighter side. (If there was hopefully!)

Then I would put 2 on each sides and hopefully see a difference again, thus eliminating the 2 lightest ones once again.

I would then be left with a pretty good idea of which one was safe or else I would just fucking guess at that point. :P

I'm tired... ;)

Actually, I think this only works with 8 pills, one of which is poisoned.

Take any six pills, put three on each side. If one side of the six weighs more, you have narrowed it down to the three pills on that side.
1) If it's even, none are poison, so weigh the last two to find the poison one. 2) If one side weighs more then take any two from that side and see which is heavier. If they're equal, it's the one you left off.

(If there are 9 pills, your inital weighing might leave one heavy side with 4 pills. There is no way to find the outlier with only 1 more weighing.)

p.s. If anyone doesn't get it, I'll post again with more details.

i think i've got it

seperate the nine pills into three groups of three. put two groups on the scale, and hold on to the last group. when you weigh it, either one side will be heavier or they have equal weights. if one side is heavier then take those three, if they weigh the same then take the three your holding. now with the remaining thee pills weigh two of them and hold on to one. if one of the side is heavier then take that safe pill, if the pills weigh the same then you are holding the safe pill!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

whoo that took forever to explain, and i better be motherfuckin right

Quote:

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Originally Posted by dackst

Actually, I think this only works with 8 pills, one of which is poisoned.

Take any six pills, put three on each side. If one side of the six weighs more, you have narrowed it down to the three pills on that side.
1) If it's even, none are poison, so weigh the last two to find the poison one. 2) If one side weighs more then take any two from that side and see which is heavier. If they're equal, it's the one you left off.

(If there are 9 pills, your inital weighing might leave one heavy side with 4 pills. There is no way to find the outlier with only 1 more weighing.)

p.s. If anyone doesn't get it, I'll post again with more details.

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dude..... it is possible with 9 pills, dont break it up into two groups of four, break it up into 3 groups of 3.

i'd agree that'd be the way to do it... i think king of the world got it

yay