19 replies to this topic

[rookie1ja]

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?

[Dredvard]

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.

[Raj777]

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

[Raj777]

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.....

[Dinghus]

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.

[lunkkun]

Spoiler for contains and proves answer

I double-checked, and rookie1jas answer is indeed the optimal answer. His answer is based on the fact that given a part of the sequence, you can determen the smallest difference you cannot make without using the last term of the sequence, and for the next term you should add that difference to the last term of your sequence, making it impossible to make a threeterm with equal sum (for example: in the sequence 0,1,2,4,7,13 the smallest difference you cannot make not using 13 is 11 (you can make the difference 10: (7+4) - (1+0) = 10), so the next term of the sequence should be 13 + 11 = 24. If you add 23 to the sequence, you can see that 13+7+4=23+1+0). This proves that rookie1jas answer (0,1,2,4,7,13,24,44) is the optimal.

[lunkkun]

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.

[Dinghus]

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.

[onehugeace]

I think the actual answer is:

Spoiler for

0, 1, 2, 4, 7, 13, 22, 23

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

[onehugeace]

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?