Determining the weight of a counterfeit coin?

Question

Bob found a pile of 50 golden dollars, along with a note stating that one of them is counterfeit and the only way to tell the weight difference is by weighing them.

1. If Bob uses a balance scale, what is the least number of times he must weigh the coins in order to determine which is the counterfeit coin?

2. Describe two different ways this could be done.

I think the answer to #1 is 2, because you can divide into two groups of 24, then, assuming they balance, weigh the remaining two to get your answer. However, I can't come up with a second way for how two weighings could be done.

Answer 1

put 10 on each side. If equal the other 30 contains the counterfeit. So do it again with the other 30
eventually, you find 10 that contain the counterfeit, and you will also know if the one that contains the counterfeit is heavier or ighter than the non-couterfeit coins.
Now repeat the process with 5 coins on each side the heavier (or lighter depending on what you determined earlier) contains the counterfeit.
Repeat the process with the final 5 coins - 2, 2 , and 1 left over. If the 2 = the two, then the left over coin is the counterfeit. If not, then the heavier (or lighter) contains the counterfeit. That leaves two coins to compare, and with the last two, the heavier, or lighter as determined earlier is the counterfiet.

Answer 2

Your solution of two groups of 24 overlooks the possibility that the counterfeit coin might be in one of these two groups of 24.
I do not know of any metal that is heavier than gold that is worth less than gold so I am going to assume the counterfeit coin is light.
Solution:
Divide the coins into 3 groups.
Weigh the two groups of 17 against each other.
If equal; the other group must contain the false coin.
Take the group that contains the false coin, split it in 3 and weigh 2 of the groups against each other. (Always round "up" when selecting the two out of the three groups.)
Continue this algorithm until the light coin is found.
I think the answer is four.
(ceil(log3(#coins)))
(ceiling is the smallest whole number larger or equal.)
(log3 is the logarithm base 3.)
If you use 4 groups there is a possibility that you could end up with 5 weighings.
If the counterfeit coin could be heavier then you could need 5 weighings and both the 3 group and 4 group algorithms are slightly different.

Answer 3

I don't think it would be 2 times. How would you know that the first groups of 24 are of equal weight and then how would you know, even if that worked, which of the 2 remaining coins was fake? Was it lighter or heavier?

Answer 4

Weigh them in water, the specific gravity will be different for the bogus coin. See "Archimedes".