SOLUTION: There were 24 coins and a balance scale. The coins are identical in every way except that one of them is counterfeit and slightly heavier than the other 23 coins. What is the least

Question 324693: There were 24 coins and a balance scale. The coins are identical in every way except that one of them is counterfeit and slightly heavier than the other 23 coins. What is the least number of weighings needed on the balance scale to determine which is the heavier coin?
Found 2 solutions by checkley77, linkbr:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
FIRST TRY 12 & 12 SELECT THE HEAVIEST SIDE
SECOND TRY 6 & 6 DITTO
THIRD TRY 3 & 3 WEIGH 2 OF THE HEAVIEST SIDE.
IF ONE WEIGHS MORE - YOU"VE FOUND IT.
IF THEY MATCH - THE THIRD COIN IS THE HEAVIEST.
THUS 4 TRIES ARE NEEDED.
Answer by linkbr(1) (Show Source): You can put this solution on YOUR website!
This can actually be solved using three weighs
1.) Divide the 24 coins into 3 stacks of 8 and weigh two.
If the two stacks that are being weighed are equal, the third stack contains the counterfeit. If they are not equal, the heavier stack contains the counterfeit.
2.) Take the stack of eight and divide it into two stacks of three and one stack of 2. Now weigh the two stacks of three. If they are the same, the stack of two contains the fraud. If the two stacks of three are not equal, the heavier one contains the deceptive coin.
3.) you now either have two coins or three coins. Either way, compare two. If they are the same, the imposter is the third coin. If they are not equal, the heavier is the counterfeit.
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