Solve 8 Coin Riddle with 2 Weighings

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To identify the one slightly heavier coin among eight identical coins using a beam balance only twice, the optimal strategy involves dividing the coins into groups. First, split the eight coins into three groups: two groups of three coins and one group of two coins. Weigh the two groups of three against each other. If one group is heavier, the heavier coin is among those three. If they balance, the heavier coin is in the group of two. Next, if the heavier group contains three coins, take any two coins from that group and weigh them against each other. If one is heavier, that is the odd coin. If they balance, the remaining coin is the heavier one. This method can also be generalized to work with nine coins or any multiple of three, allowing for the identification of the heavier coin with the same number of weighings.
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You are given eight coins of same denomination, out of the eight coins one coin is very very slightly heavy than the remaining seven such that you cannot separate it by weighing it in your palms. You are given a beam balance with the help of which you have to separate out the very slightly heavy coin but you are allowed to use the balance only twice.
 
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There's probably a better way to word this but here is my solution.

Put 3 on one side, 3 on the other. If the scale doesn't tip, weigh the 2 that were not on the scale.

If one of the 3 tips, pick 2 of those and put them on the scale, if it balances it's the 3rd one of that group if not the scale will tip.
 
Charmar said:
There's probably a better way to word this but here is my solution.

Put 3 on one side, 3 on the other. If the scale doesn't tip, weigh the 2 that were not on the scale.

If one of the 3 tips, pick 2 of those and put them on the scale, if it balances it's the 3rd one of that group if not the scale will tip.

absolutely correct :approve:
 
You can do the same with 9 coins.
Or, in general, with 3n coins if you can use the balance n times.
 
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