rookie1ja 15 Posted March 30, 2007 Report Share Posted March 30, 2007 Weighing III. - Back to the Water and Weighing Puzzles This puzzle goes a step further from the previous one. You have eight bags, each of them containing 48 coins. Five of those bags contain only true coins, the rest of them contain fake coins. A fake coin weighs 1 gram less than a real coin. You have an accurate scale, with the precision of up to 1 gram. Weighing only once and using the minimum number of coins, how can you find the bags containing the fake coins? This old topic is locked since it was answered many times. You can check solution in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers. Weighing III. - solution Similar to the former brain teaser. I take out 0 (no coin from the first bag), 1 (one coin from the second bag etc.), 2, 4, 7, 13, 24, 44 coins (from the last 8th bag). Each triple is unique enabling an easy way to identify the bags with fake coins (using only 95 coins). This puzzle is a step further than the previous one (weight of fake and non fake coins is the same as the weight of bears in the previous puzzle). You have eight bags, each of them containing 48 coins. Five of these bags contain only true coins, the rest of them contain fake coins. Fake coins weigh 1 gram less than the real coins. You do not know what bags have fake coins and what bags have real coins. You can use a scale, a dynamometer type one, with precision up to 1 gram (an accurate weighing machine). Weighing only once and using the minimum number of coins, how can you find the bags containing the fake coins? Link to post Share on other sites

Guest Posted August 12, 2007 Report Share Posted August 12, 2007 Since the scale only measures to 1 gram, is it logical to assume you can remove coins from the bags to measure them? And what do you mean by you can only measure once? Measure once per bag, or once in total? Fake coin bags should weigh 48 grams less than the real bags. Link to post Share on other sites

Guest Posted August 12, 2007 Report Share Posted August 12, 2007 Since the scale only measures to 1 gram, is it logical to assume you can remove coins from the bags to measure them? Yes, it logical to assume you can remove coins from the bags. No, the scale can not only measure up to 1 gram; it is accurate within 1 gram. Assume that it can handle weighing all coins at once. And what do you mean by you can only measure once? Measure once per bag, or once in total? Once in total. Fake coin bags should weigh 48 grams less than the real bags. Correct. Link to post Share on other sites

Guest Posted November 21, 2007 Report Share Posted November 21, 2007 I think I found the solution: Assign numbers to each of the bags such that the sum of any 3 numbers assigned to the bags does not equal the sum of any other three bags. Take the number of coins that corresponds to the number on the bag. Put them all in an individual pile on the scale and push the button and record the number on the scale. With simple math you can find which bags' coins are causing the difference in weight compared to what it would be if all the coins were real. For instance: Bag 1: 0 Bag 2: 1 Bag 3: 2* Bag 4: 3 Bag 5: 7* Bag 6: 13 Bag 7: 24 Bag 8: 45* In this scenario, if bags 3, 5, and 8 are fake, there would be a difference of 54 grams registered on the scale. We would know that only 45, 7, and 2 add together to equal 54, so it must be bags 3, 5, and 8. I don't know if this is the minimum, but I think it is. Link to post Share on other sites

Guest Posted December 19, 2007 Report Share Posted December 19, 2007 soldier3001, your reasoning is sound, but your solution is (barely) suboptimal. The original answer is the best answer, and differs from your solution only by the last number of coins being 44 instead of 45. Link to post Share on other sites

Guest Posted December 21, 2007 Report Share Posted December 21, 2007 well why not choose the solultion as below 0, 1, 2, 3, 6, 11, 20, 37 each triple is different Link to post Share on other sites

Guest Posted December 21, 2007 Report Share Posted December 21, 2007 the solution I provided above is wrong the correct answer is 0, 1. 2, 4, 7, 13, 24 Link to post Share on other sites

Guest Posted December 21, 2007 Report Share Posted December 21, 2007 the solution I provided above is wrong the correct answer is 0, 1. 2, 4, 7, 13, 24 Yes, except there are eight bags and you've only taken from seven (actually from six, taking 0 from two bags). Link to post Share on other sites

Guest Posted February 22, 2008 Report Share Posted February 22, 2008 the solution I provided above is wrong the correct answer is 0, 1. 2, 4, 7, 13, 24 Yes, except there are eight bags and you've only taken from seven (actually from six, taking 0 from two bags). I wan to understand how taking zero coins from two bags will let you identity 3 bags containing fake coins? Eg: If the total weight of the coins taken as you have mentioned, is less than the actual by 20gms. Then the solution is (13+7+(either bag1 or bag8, both of which you have omitted)). So how will you identify between bag1 and bag8? Link to post Share on other sites

Guest Posted February 22, 2008 Report Share Posted February 22, 2008 (edited) The solution : 1 2 3 4 7 12 21 38. The total coins taken for weighing will be only 88. Edited February 22, 2008 by rahul Link to post Share on other sites

rookie1ja 15 Posted February 22, 2008 Author Report Share Posted February 22, 2008 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? Link to post Share on other sites

Guest Posted May 4, 2008 Report Share Posted May 4, 2008 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. Link to post Share on other sites

Guest Posted May 19, 2008 Report Share Posted May 19, 2008 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 Link to post Share on other sites

Guest Posted May 19, 2008 Report Share Posted May 19, 2008 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..... Link to post Share on other sites

Guest Posted September 2, 2008 Report Share Posted September 2, 2008 (edited) 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 September 2, 2008 by Dinghus Link to post Share on other sites

Guest Posted September 22, 2008 Report Share Posted September 22, 2008 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. Link to post Share on other sites

Guest Posted September 22, 2008 Report Share Posted September 22, 2008 (edited) 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 September 22, 2008 by lunkkun Link to post Share on other sites

Guest Posted September 22, 2008 Report Share Posted September 22, 2008 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. Link to post Share on other sites

Guest Posted February 6, 2010 Report Share Posted February 6, 2010 I think the actual answer is: 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. Link to post Share on other sites

Guest Posted February 6, 2010 Report Share Posted February 6, 2010 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? Link to post Share on other sites

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