Solve 8 Coin Riddle with 2 Weighings

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Discussion Overview

The discussion revolves around a riddle involving eight coins, where one coin is slightly heavier than the others. Participants explore methods to identify the heavier coin using a beam balance with only two weighings. The scope includes problem-solving and mathematical reasoning.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant presents a solution to identify the heavier coin using two weighings.
  • Another participant expresses agreement with the proposed solution, indicating it is correct.
  • A different participant suggests that the same method can be applied to nine coins or generally to 3n coins with n weighings.

Areas of Agreement / Disagreement

There is some agreement on the validity of the proposed solution, but the discussion introduces the possibility of extending the method to a larger number of coins, indicating multiple perspectives on the problem.

Contextual Notes

The discussion does not clarify the specific methods used in the proposed solutions or the reasoning behind them, leaving some assumptions and steps unresolved.

mia5
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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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