rookie1ja Posted March 30, 2007 Report Share Posted March 30, 2007 Weighing II. - Back to the Water and Weighing Puzzles A genuine gummy drop bear has a mass of 10 grams, while an imitation gummy drop bear has a mass of 9 grams. Spike has 7 cartons of gummy drop bears, 4 of which contain real gummy drop bears and the others – imitation. Using a scale only once and the minimum number of gummy drop bears, how can Spike determine which cartons contain real gummy drop bears? 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 II. - solution Spike uses 51 gummy drop bears: from the 7 boxes he takes respectively 0, 1, 2, 4, 7, 13, and 24 bears. The notion is that each box of imitation bears will subtract its number of bears from the total "ideal" weight of 510 grams (1 gram of missing weight per bear), so Spike weighs the bears, subtracts the result from 510 to obtain a number N, and finds the unique combination of 3 numbers from the above list (since there are 3 "imitation" boxes) that sum to N. The trick is for the sums of all triples selected from the set S of numbers of bears to be unique. To accomplish this, I put numbers into S one at a time in ascending order, starting with the obvious choice, 0. (Why is this obvious? If I'd started with k > 0, then I could have improved on the resulting solution by subtracting k from each number) Each new number obviously had to be greater than any previous, because otherwise sums are not unique, but also the sums it made when paired with any previous number had to be distinct from all previous pairs (otherwise when this pair is combined with a third number you can't distinguish it from the other pair)--except for the last box, where we can ignore this point. And most obviously all the new triples had to be distinct from any old triples; it was easy to find what the new triples were by adding the newest number to each old sum of pairs. Now, in case you're curious, the possible weight deficits and their unique decompositions are: 3 = 0 + 1 + 2 5 = 0 + 1 + 4 6 = 0 + 2 + 4 7 = 1 + 2 + 4 8 = 0 + 1 + 7 9 = 0 + 2 + 7 10 = 1 + 2 + 7 11 = 0 + 4 + 7 12 = 1 + 4 + 7 13 = 2 + 4 + 7 14 = 0 + 1 + 13 15 = 0 + 2 + 13 16 = 1 + 2 + 13 17 = 0 + 4 + 13 18 = 1 + 4 + 13 19 = 2 + 4 + 13 20 = 0 + 7 + 13 21 = 1 + 7 + 13 22 = 2 + 7 + 13 24 = 4 + 7 + 13 25 = 0 + 1 + 24 26 = 0 + 2 + 24 27 = 1 + 2 + 24 28 = 0 + 4 + 24 29 = 1 + 4 + 24 30 = 2 + 4 + 24 31 = 0 + 7 + 24 32 = 1 + 7 + 24 33 = 2 + 7 + 24 35 = 4 + 7 + 24 37 = 0 + 13 + 24 38 = 1 + 13 + 24 39 = 2 + 13 + 24 41 = 4 + 13 + 24 44 = 7 + 13 + 24 Note that there had to be (7 choose 3) distinct values; they end up ranging from 3 to 44 inclusive with 7 numbers missing: 4, 23, 34, 36, 40, 42, and 43. Real gummy drop bears have a mass of 10 grams, while imitation gummy drop bears have a mass of 9 grams. Spike has 7 cartons of gummy drop bears, 4 of which contain real gummy drop bears, the others imitation. Using a scale only once and the minimum number of gummy drop bears, how can Spike determine which cartons contain real gummy drop bears? Edit: scales (for this case) = weighing machine (like in science labs, so no pair of scales) and the plates are big enough for all possible weighings Edit: each carton contains 50 bears 1 Link to comment Share on other sites More sharing options...

Guest Posted April 20, 2007 Report Share Posted April 20, 2007 This sequence uses minimum gummy bears. 0, 1, 2, 3, 6, 11, 20 This covers all unique sums from 3 to 37 ( 35 unique values = 7 choose 3). Link to comment Share on other sites More sharing options...

Guest Posted April 20, 2007 Report Share Posted April 20, 2007 Sorry, the previous sequence posted by me does not work. You were right. Link to comment Share on other sites More sharing options...

Guest Posted July 22, 2007 Report Share Posted July 22, 2007 Each new number obviously had to be greater than any previous, because otherwise sums are not unique, but also the sums it made when paired with any previous number had to be distinct from all previous pairs (otherwise when this pair is combined with a third number you can't distinguish it from the other pair)--except for the last box, where we can ignore this point. While the solution is correct, the logic quoted above in attaining it is not. By that logic the 6th bag (13) could be 10 instead as no two numbers before it add up to 10. However, 10+1 = 7+4. Therefore, you also have to check to be sure that the number plus any previous number does not sum to the same as any two previous numbers. That will get you a little farther (to 13 appropriately). But even with that additional caveat you'll find problems. 21 would be the next available number using those rules, but 0+1+21 = 2+7+13. So you'd also have to add the restriction that the new number plus any 2 previous numbers can't sum to the sum of any three previous numbers, which is just stating the problem of unique triples over again. So... here's the logic you have to use to solve it: Start with 0, 1, 2, 4 and thereafter the next number in the series = the sum of the last three numbers. The sum of the last three numbers in the series is the highest possible sum to that point, and the new number must at a minimum add 0 and 1, thereby guaranteeing unique triples. Link to comment Share on other sites More sharing options...

Guest Posted October 9, 2007 Report Share Posted October 9, 2007 I have one complaint about this puzzle - The type of scale was not defined in the puzzle. Typically the type of scale used in these weight/weighing puzzles is a balance, where some amount of billiard balls are weighed against another amount of billiard balls. However the solution to this puzzle, points to a scale that is not a simple balance. For this puzzle, the scale has to be either a balance with weights, or your typical scale used in science labs. Other than that small vague area of the scale, its a perfectly delightful puzzle that requires great ingenuity to solve. Bravo. * Totally nitpicking on the size of the plate, we all know these puzzles deal in ideal (unrealistic) circumstances. Like physics problems that you get before learning how to account for friction. However... the scale would need a rather large weighing plate to allow for 6 sets (the seventh has no bears at all) to be placed on it without getting the sets all mixed together. If spike had to use little plastic baggies to keep the bears seperate, then he would be forced to weigh the the empty baggies first, thus using the scale twice. Link to comment Share on other sites More sharing options...

rookie1ja Posted October 9, 2007 Author Report Share Posted October 9, 2007 I have one complaint about this puzzle - The type of scale was not defined in the puzzle. Typically the type of scale used in these weight/weighing puzzles is a balance, where some amount of billiard balls are weighed against another amount of billiard balls. However the solution to this puzzle, points to a scale that is not a simple balance. For this puzzle, the scale has to be either a balance with weights, or your typical scale used in science labs. right, I have edited the puzzle Link to comment Share on other sites More sharing options...

Guest Posted November 12, 2007 Report Share Posted November 12, 2007 to be placed on it without getting the sets all mixed together. If spike had to use little plastic baggies to keep the bears seperate, then he would be forced to weigh the the empty baggies first, thus using the scale twice. the beauty is they can mix you can eat them afterwards if you want Link to comment Share on other sites More sharing options...

Guest Posted December 10, 2007 Report Share Posted December 10, 2007 The trick is for the sums of all triples selected from the set S of numbers of bears to be unique. could you please elaborate on this line Link to comment Share on other sites More sharing options...

Guest Posted December 18, 2007 Report Share Posted December 18, 2007 Weighing II. - Back to the Water and Weighing Puzzles Real gummy drop bears have a mass of 10 grams, while imitation gummy drop bears have a mass of 9 grams. Spike has 7 cartons of gummy drop bears, 4 of which contain real gummy drop bears, the others imitation. Using a scale only once and the minimum number of gummy drop bears, how can Spike determine which cartons contain real gummy drop bears? Edit: scales (for this case) = weighing machine (like in science labs, so no pair of scales) and the plates are big enough for all possible weighings Weighing II. - solution Spike uses 51 gummy drop bears: from the 7 boxes he takes respectively 0, 1, 2, 4, 7, 13, and 24 bears. The notion is that each box of imitation bears will subtract its number of bears from the total "ideal" weight of 510 grams (1 gram of missing weight per bear), so Spike weighs the bears, subtracts the result from 510 to obtain a number N, and finds the unique combination of 3 numbers from the above list (since there are 3 "imitation" boxes) that sum to N. The trick is for the sums of all triples selected from the set S of numbers of bears to be unique. To accomplish this, I put numbers into S one at a time in ascending order, starting with the obvious choice, 0. (Why is this obvious? If I'd started with k > 0, then I could have improved on the resulting solution by subtracting k from each number) Each new number obviously had to be greater than any previous, because otherwise sums are not unique, but also the sums it made when paired with any previous number had to be distinct from all previous pairs (otherwise when this pair is combined with a third number you can't distinguish it from the other pair)--except for the last box, where we can ignore this point. And most obviously all the new triples had to be distinct from any old triples; it was easy to find what the new triples were by adding the newest number to each old sum of pairs. Now, in case you're curious, the possible weight deficits and their unique decompositions are: 3 = 0 + 1 + 2 5 = 0 + 1 + 4 6 = 0 + 2 + 4 7 = 1 + 2 + 4 8 = 0 + 1 + 7 9 = 0 + 2 + 7 10 = 1 + 2 + 7 11 = 0 + 4 + 7 12 = 1 + 4 + 7 13 = 2 + 4 + 7 14 = 0 + 1 + 13 15 = 0 + 2 + 13 16 = 1 + 2 + 13 17 = 0 + 4 + 13 18 = 1 + 4 + 13 19 = 2 + 4 + 13 20 = 0 + 7 + 13 21 = 1 + 7 + 13 22 = 2 + 7 + 13 24 = 4 + 7 + 13 25 = 0 + 1 + 24 26 = 0 + 2 + 24 27 = 1 + 2 + 24 28 = 0 + 4 + 24 29 = 1 + 4 + 24 30 = 2 + 4 + 24 31 = 0 + 7 + 24 32 = 1 + 7 + 24 33 = 2 + 7 + 24 35 = 4 + 7 + 24 37 = 0 + 13 + 24 38 = 1 + 13 + 24 39 = 2 + 13 + 24 41 = 4 + 13 + 24 44 = 7 + 13 + 24 that there had to be (7 choose 3) distinct values; they end up ranging from 3 to 44 inclusive with 7 numbers missing: 4, 23, 34, 36, 40, 42, and 43. This answer gives a total required of 0 + 1 + 2 + 4 + 7 + 13 + 24 = 51 bears. This is not a minimum, however, so it is wrong. Since we know that there are exactly four of the seven boxes that contain fake bears, we can use this fact to get a better number sequence. 0 1 2 3 4 8 15 for a total of 33 bears. Any group of four of these numbers yields, I believe, a unique sum. Link to comment Share on other sites More sharing options...

rookie1ja Posted December 18, 2007 Author Report Share Posted December 18, 2007 Weighing II. - Back to the Water and Weighing Puzzles Real gummy drop bears have a mass of 10 grams, while imitation gummy drop bears have a mass of 9 grams. Spike has 7 cartons of gummy drop bears, 4 of which contain real gummy drop bears, the others imitation. Using a scale only once and the minimum number of gummy drop bears, how can Spike determine which cartons contain real gummy drop bears? Edit: scales (for this case) = weighing machine (like in science labs, so no pair of scales) and the plates are big enough for all possible weighings Weighing II. - solution Spike uses 51 gummy drop bears: from the 7 boxes he takes respectively 0, 1, 2, 4, 7, 13, and 24 bears. The notion is that each box of imitation bears will subtract its number of bears from the total "ideal" weight of 510 grams (1 gram of missing weight per bear), so Spike weighs the bears, subtracts the result from 510 to obtain a number N, and finds the unique combination of 3 numbers from the above list (since there are 3 "imitation" boxes) that sum to N. The trick is for the sums of all triples selected from the set S of numbers of bears to be unique. To accomplish this, I put numbers into S one at a time in ascending order, starting with the obvious choice, 0. (Why is this obvious? If I'd started with k > 0, then I could have improved on the resulting solution by subtracting k from each number) Each new number obviously had to be greater than any previous, because otherwise sums are not unique, but also the sums it made when paired with any previous number had to be distinct from all previous pairs (otherwise when this pair is combined with a third number you can't distinguish it from the other pair)--except for the last box, where we can ignore this point. And most obviously all the new triples had to be distinct from any old triples; it was easy to find what the new triples were by adding the newest number to each old sum of pairs. Now, in case you're curious, the possible weight deficits and their unique decompositions are: 3 = 0 + 1 + 2 5 = 0 + 1 + 4 6 = 0 + 2 + 4 7 = 1 + 2 + 4 8 = 0 + 1 + 7 9 = 0 + 2 + 7 10 = 1 + 2 + 7 11 = 0 + 4 + 7 12 = 1 + 4 + 7 13 = 2 + 4 + 7 14 = 0 + 1 + 13 15 = 0 + 2 + 13 16 = 1 + 2 + 13 17 = 0 + 4 + 13 18 = 1 + 4 + 13 19 = 2 + 4 + 13 20 = 0 + 7 + 13 21 = 1 + 7 + 13 22 = 2 + 7 + 13 24 = 4 + 7 + 13 25 = 0 + 1 + 24 26 = 0 + 2 + 24 27 = 1 + 2 + 24 28 = 0 + 4 + 24 29 = 1 + 4 + 24 30 = 2 + 4 + 24 31 = 0 + 7 + 24 32 = 1 + 7 + 24 33 = 2 + 7 + 24 35 = 4 + 7 + 24 37 = 0 + 13 + 24 38 = 1 + 13 + 24 39 = 2 + 13 + 24 41 = 4 + 13 + 24 44 = 7 + 13 + 24 that there had to be (7 choose 3) distinct values; they end up ranging from 3 to 44 inclusive with 7 numbers missing: 4, 23, 34, 36, 40, 42, and 43. This answer gives a total required of 0 + 1 + 2 + 4 + 7 + 13 + 24 = 51 bears. This is not a minimum, however, so it is wrong. Since we know that there are exactly four of the seven boxes that contain fake bears, we can use this fact to get a better number sequence. 0 1 2 3 4 8 15 for a total of 33 bears. Any group of four of these numbers yields, I believe, a unique sum. I don't think so ... what about 15 8+4+2+1=15 8+4+3+0=15 Link to comment Share on other sites More sharing options...

Guest Posted December 19, 2007 Report Share Posted December 19, 2007 I don't think so ... what about 15 8+4+2+1=15 8+4+3+0=15 You're right, of course, rookie. The original answer was indeed the optimal solution. Sorry. Link to comment Share on other sites More sharing options...

rookie1ja Posted December 20, 2007 Author Report Share Posted December 20, 2007 I don't think so ... what about 15 8+4+2+1=15 8+4+3+0=15 You're right, of course, rookie. The original answer was indeed the optimal solution. Sorry. no worries ... if you know some other good brain teasers, you can post it in the new puzzles section Link to comment Share on other sites More sharing options...

Guest Posted February 22, 2008 Report Share Posted February 22, 2008 Another solution: spike stands on the scale, holding all the boxes. He drops on box at a time, and sees the difference in weight change for each individual box, and can infer which box contains the real gummy bears, and which contain fake ones. Link to comment Share on other sites More sharing options...

Guest Posted April 4, 2008 Report Share Posted April 4, 2008 correct me if im wrong but, do you not need to be told how many each carton contains, what if a carton only contains 20 gummy drop bears? your solution would not be correct - but if you were to asume that it has in excess of this, as you have then you could be correct. But yeh, would'nt you need to be a bit more specific? Link to comment Share on other sites More sharing options...

rookie1ja Posted April 4, 2008 Author Report Share Posted April 4, 2008 correct me if im wrong but, do you not need to be told how many each carton contains, what if a carton only contains 20 gummy drop bears? your solution would not be correct - but if you were to asume that it has in excess of this, as you have then you could be correct. But yeh, would'nt you need to be a bit more specific? good point Edit: each carton contains 50 bears Link to comment Share on other sites More sharing options...

Guest Posted April 30, 2008 Report Share Posted April 30, 2008 Weighing II. - Back to the Water and Weighing Puzzles Real gummy drop bears have a mass of 10 grams, while imitation gummy drop bears have a mass of 9 grams. Spike has 7 cartons of gummy drop bears, 4 of which contain real gummy drop bears, the others imitation. Using a scale only once and the minimum number of gummy drop bears, how can Spike determine which cartons contain real gummy drop bears? Edit: scales (for this case) = weighing machine (like in science labs, so no pair of scales) and the plates are big enough for all possible weighings Edit: each carton contains 50 bears Weighing II. - solution Spike uses 51 gummy drop bears: from the 7 boxes he takes respectively 0, 1, 2, 4, 7, 13, and 24 bears. The notion is that each box of imitation bears will subtract its number of bears from the total "ideal" weight of 510 grams (1 gram of missing weight per bear), so Spike weighs the bears, subtracts the result from 510 to obtain a number N, and finds the unique combination of 3 numbers from the above list (since there are 3 "imitation" boxes) that sum to N. The trick is for the sums of all triples selected from the set S of numbers of bears to be unique. To accomplish this, I put numbers into S one at a time in ascending order, starting with the obvious choice, 0. (Why is this obvious? If I'd started with k > 0, then I could have improved on the resulting solution by subtracting k from each number) Each new number obviously had to be greater than any previous, because otherwise sums are not unique, but also the sums it made when paired with any previous number had to be distinct from all previous pairs (otherwise when this pair is combined with a third number you can't distinguish it from the other pair)--except for the last box, where we can ignore this point. And most obviously all the new triples had to be distinct from any old triples; it was easy to find what the new triples were by adding the newest number to each old sum of pairs. Now, in case you're curious, the possible weight deficits and their unique decompositions are: 3 = 0 + 1 + 2 5 = 0 + 1 + 4 6 = 0 + 2 + 4 7 = 1 + 2 + 4 8 = 0 + 1 + 7 9 = 0 + 2 + 7 10 = 1 + 2 + 7 11 = 0 + 4 + 7 12 = 1 + 4 + 7 13 = 2 + 4 + 7 14 = 0 + 1 + 13 15 = 0 + 2 + 13 16 = 1 + 2 + 13 17 = 0 + 4 + 13 18 = 1 + 4 + 13 19 = 2 + 4 + 13 20 = 0 + 7 + 13 21 = 1 + 7 + 13 22 = 2 + 7 + 13 24 = 4 + 7 + 13 25 = 0 + 1 + 24 26 = 0 + 2 + 24 27 = 1 + 2 + 24 28 = 0 + 4 + 24 29 = 1 + 4 + 24 30 = 2 + 4 + 24 31 = 0 + 7 + 24 32 = 1 + 7 + 24 33 = 2 + 7 + 24 35 = 4 + 7 + 24 37 = 0 + 13 + 24 38 = 1 + 13 + 24 39 = 2 + 13 + 24 41 = 4 + 13 + 24 44 = 7 + 13 + 24 Note that there had to be (7 choose 3) distinct values; they end up ranging from 3 to 44 inclusive with 7 numbers missing: 4, 23, 34, 36, 40, 42, and 43. Excellent puzzle. =8o) Although in my google gadget where I get a feed of these things, it does not pick up your edits, and I was puzzling away for a good while before "giving up" and going to the BrainDen page -- only to find the addendums. hehe Link to comment Share on other sites More sharing options...

Guest Posted July 4, 2008 Report Share Posted July 4, 2008 Is this a possible solution? 1 2 3 4 7 12 21 <-- 50 gummies so you take the number of grams below 50 x 10 = 500 if the difference is 6 -> carton 3 and 2 of the cartons to carton 3's left are immitations 7-9 -> carton 4 and 2 of the cartons to carton 3's left are immitations 10-14 -> carton 5 and 2 of the cartons to carton 3's left are immitations 15-23 -> carton 6 and 2 of the cartons to carton 3's left are immitations 24-40 -> carton 7 and 2 of the cartons to carton 3's left are immitations and since the difference would be a unique sum of 3 of the numbers in the sequence, it should be easy to find the remaining 2 immitations, that is, assuming my sequence is correct. Link to comment Share on other sites More sharing options...

Guest Posted September 16, 2008 Report Share Posted September 16, 2008 Is this a possible solution? 1 2 3 4 7 12 21 <-- 50 gummies so you take the number of grams below 50 x 10 = 500 if the difference is 6 -> carton 3 and 2 of the cartons to carton 3's left are immitations 7-9 -> carton 4 and 2 of the cartons to carton 4's left are immitations 10-14 -> carton 5 and 2 of the cartons to carton 5's left are immitations 15-23 -> carton 6 and 2 of the cartons to carton 6's left are immitations 24-40 -> carton 7 and 2 of the cartons to carton 7's left are immitations and since the difference would be a unique sum of 3 of the numbers in the sequence, it should be easy to find the remaining 2 immitations, that is, assuming my sequence is correct. I don't think you meant to put "carton 3's left" over and over so I changed it above but anyway: No it is not a solution because 2 + 3 = 1 + 4. For example if you find the difference to be twelve grams that means carton 5 and two to its left. The two to the left must add up to 12 - 7 = 5 and that could mean carton 1 and carton 4 or carton 2 and carton 3. Link to comment Share on other sites More sharing options...

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