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Weighing III.


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19 replies to this topic

#11 rookie1ja

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Posted 22 February 2008 - 11:27 AM

The solution : 1 2 3 4 7 12 21 38. The total coins taken for weighing will be only 88.

what about a difference of 12 ... is it 1+4+7 or 2+3+7?
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#12 Dredvard

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Posted 04 May 2008 - 02:08 AM

rookie, you seem to have an inherant understanding of this puzzle. How do you know these are unique triples and that its the optimal solution? This one blew me away... and I've posed it to a bunch of people and they've all struggled with it too.
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#13 Raj777

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Posted 19 May 2008 - 01:19 PM

I think the solution is 28 coins

0 from 1st bag
1 from 2nd bag and so forth 0+1+2+3+4+5+6+7 = 28 coins

The bags can be identified based on the difference between the total weight and the weight if all coins were true.... This would be unique
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#14 Raj777

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Posted 19 May 2008 - 03:11 PM

I think the solution is 28 coins

0 from 1st bag
1 from 2nd bag and so forth 0+1+2+3+4+5+6+7 = 28 coins

The bags can be identified based on the difference between the total weight and the weight if all coins were true.... This would be unique



Sorry was wrong!!!!
Should have been

1 from 1st bag, 2 from 2nd bag and so forth, so we need 36 coins.....
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#15 Dinghus

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Posted 02 September 2008 - 07:11 PM

Wow. Everybody is wrong. Amazing.

What is one of the conditions here? That the scale is accurate to within 1 gram. Since the fakes are 1 gram less than the trues, if you measure 1 coin you will not be able to determine if it is fake or not since 1 gram is within the accuracy of the scale.

Say a single coin weighs 5 grams. A fake coin ways 4 grams. But a real coin COULD weigh in anywhere from 4 to 6 grams. Thus you must have MORE than 1 coin from each bag to determine if it is fake or not.

Continuing on that line. In reality it would take more than 1 weighing because you would have to first determine what a REAL coin weighs according to that scale. Once that is done, you can work out the rest of it.

Everybody back to the drawing boards and come back with a correct solution.

Edited by Dinghus, 02 September 2008 - 07:18 PM.

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#16 lunkkun

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Posted 22 September 2008 - 05:27 PM

Spoiler for contains and proves answer

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#17 lunkkun

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Posted 22 September 2008 - 05:33 PM


I think the solution is 28 coins

0 from 1st bag
1 from 2nd bag and so forth 0+1+2+3+4+5+6+7 = 28 coins

The bags can be identified based on the difference between the total weight and the weight if all coins were true.... This would be unique

Sorry was wrong!!!!
Should have been

1 from 1st bag, 2 from 2nd bag and so forth, so we need 36 coins.....

In both answers, if the outcome differs 10 grams, you don't know if it was 1+4+5 or 2+3+5.

Edited by lunkkun, 22 September 2008 - 05:34 PM.

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#18 Dinghus

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Posted 22 September 2008 - 06:40 PM

It will actually take a MINIMUM of 4 weighings to come up with an answer.

You FIRST have to weigh 3 coins minimum to determine the weight of either a fake or a real coin on the scale provided.

The best thing for this problem is to say the scale is 100% accurate.
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#19 onehugeace

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Posted 06 February 2010 - 02:31 AM

I think the actual answer is:

Spoiler for


This is a total of 72 coins, and any 3 numbers produces a unique sum that is not matched by any other three numbers.
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#20 onehugeace

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Posted 06 February 2010 - 02:35 AM

It will actually take a MINIMUM of 4 weighings to come up with an answer.

You FIRST have to weigh 3 coins minimum to determine the weight of either a fake or a real coin on the scale provided.

The best thing for this problem is to say the scale is 100% accurate.


I don't think you need to weigh more than once, because the weight of any amount of coins, if all coins were true coins, will always be a multiple of 8. Knowing that three bags are filled with fake coins means that you should be able to figure out the total number of fake coins in the pile that is being weighed, because the true weight should be that multiple of 8. Therefore, it doesn't matter whether you know the weight of a true coin or not. Can you think of a counterexample?
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