Well, since it doesn't say that you have to tell which pill is the poison one, this problem should only take two weighings at most, usually it would only take one.
Grab two random pills, weigh them against each other, if they come out even,swallow one of those two pills.
If they don't come out even then just swallow one of the ten pills that weren't weighed
Monday, February 24, 2003
All of the normal pills are poison, so you are trying to find the one that doesn't way the same. But you are right, it can be done in at most two weighings - the way I was told the riddle a long time ago, though, is that you have to find which pill weighs different and whether it is heavier or lighter. This takes 3 different balances.
Friday, March 07, 2003
The solution for this problem is far to big.
There's a far easier way
3 on each side of the balance
6 pills set aside
--> you know in which group of 6 the different pill is
3 of the normal pills (the pills which don't contain the different one) on one side and 3 of the group with the different in it on the other side.
--> you know in which group of 3 the different pill is
1 pil on each side
1 you set aside
--> you survive
Well unless you mix up the piles of pills this should work well:)
If i made a flaw or did something stupid please tell me by mail or let me know @ my website
Thierry Schellenbach http://www.top-download.net
Saturday, March 12, 2005
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