Arthur D. answered 09/03/14
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Mathematics Tutor With a Master's Degree In Mathematics
If I know that the poisoned gummy bear weighs less than all the others I can solve the problem, or if I know that the poisoned gummy bear weighs more than all the others I can solve the problem also. I will assume that the gummy bear that is poison weighs more than all the others. The solution is the same if the gummy bear weighs less than all the others.
put 3 gummy bears on one side and 3 gummy bears on the other side of the scale
if they weigh the same, the poisoned one is with the other 6 (1 weighing)
now put the other 6 on the scale, 3 on one side and 3 on the other
the side of the scale that goes down contains the heavier poisoned gummy bear (2 weighings)
you know that one of the 3 gummy bears on the side that went down is poisoned
take any 2 and weigh them
if the scale balances, then the one left over is poison
if the scale doesn't balance the side that goes down contains the poisoned gummy bear (3 weighings)
if on the first weighing, the scale doesn't balance, you know the poisoned gummy bear is on the side that goes down
you then take any 2 and weigh them and I have already explained what to do in this case