This week's giveaway is in the EJB and other Java EE Technologies forum. We're giving away four copies of EJB 3 in Action and have Debu Panda, Reza Rahman, Ryan Cuprak, and Michael Remijan on-line! See this thread for details.

This is also pretty easy, but I do not like puzzles I have to spend hours working on There are 9 coins and one of them is false - it weight less than rest. You have a balance. How to find false coin with only two balancings? (veterans of triangle counting and rope burning are not welcome )

Test six coins - 3 on each side of the balance. If these two groups of three are balanced, then remove all coins and only consider the three unchecked coins; otherwise, consider the lighter group of coins weighed. Considering these three coins, place one on each side of the balance. If these two coins are balanced, then the one remaining unweighed coin is the false coin; otherwise, the coin that proved lighter on the balance is the false coin.

Apparently people in this forum have significant experience with false coins. :roll: Kinda scaring... Michael, you do not need to know the answer to whisper something in my ear

Dirk Schreckmann
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Originally posted by Mapraputa Is: Apparently people in this forum have significant experience with false coins.

I recently returned from a trip to El Salvador. I brought back with me a handful of coins, that while worth only about 3 cents each, are reported to work in the local parking meters as quarters - I wouldn't know anything about that. And yes, they are lighter than quarters.

make the three stack of 3 coins each 1 1 1 1 1 1 1 1 1 weigh any two stacks 1 1 1 1 1 1 either both weighs equal or one of the stack weighs less if both weighs equal then the wrong coin must be in the third stack 1 1 1 so take out one coin and weigth the rest of the two so either of the coin which weighs less is the wrong one and if both are equal then the coin which you have taken out is the wrong one if both does not weighs equal then the wrong coin must be in the stack that wighs less, take that stack which weighs less and take out one coin and weigth the rest of the two so either of the coin which weighs less is the wrong one and if both are equal then the coin which you have taken out is the wrong one