How to Determine Counterfeit Coins with Only One Weighing

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SUMMARY

To determine counterfeit coins among seven stacks, each containing 100 coins, only one weighing is necessary using an analytical scale accurate to 0.1 grams. Real coins weigh 10 grams, while counterfeit coins weigh 11 grams. By taking a specific number of coins from each stack and weighing them collectively, the total weight will indicate the number of counterfeit stacks based on the deviation from the expected weight. This method efficiently identifies both the quantity and the specific stacks of counterfeit coins.

PREREQUISITES
  • Understanding of basic arithmetic and weight measurement.
  • Familiarity with analytical scales and their precision.
  • Knowledge of logical deduction and problem-solving techniques.
  • Concept of weight deviation and its implications in identifying counterfeit items.
NEXT STEPS
  • Research the principles of weight measurement and accuracy in analytical scales.
  • Explore logical deduction strategies for problem-solving in mathematics.
  • Learn about counterfeit detection methods in various industries.
  • Investigate similar puzzles involving weighing and logical reasoning.
USEFUL FOR

This discussion is beneficial for mathematicians, puzzle enthusiasts, educators teaching logical reasoning, and anyone interested in problem-solving strategies related to weight measurement and counterfeit detection.

AlexChandler
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There are seven stacks of coins, each consisting of 100 coins.
You know that real coins weigh 10 grams, but counterfeit coins weigh 11 grams.
You have an analytical scale that is accurate to .1 gram.
Each stack consists of either all real coins or all counterfeit coins, none of the stacks are mixed.
You do not know how many stacks are counterfeit or even if any of them are.
You want to determine how many stacks are counterfeit (if any), and precisely which ones are.
You want to do this in the least possible amount of weighings.

Whats the fewest number of weighings you need to make?
And how do you do it?
 
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Is there a catch? I make it 1.

Take 1, 2, 4, 8, 16, 32, 64 from each pile, measure the excess weight and convert it to a binary number
 
Jonathan Scott said:
Is there a catch? I make it 1.

Take 1, 2, 4, 8, 16, 32, 64 from each pile, measure the excess weight and convert it to a binary number

I'm impressed. :-p That was fast!
 

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