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How do I solve the coin problem?

Math Logic Problem - Please HELP!!!?

  • You work for a company that develops puzzles for the general public. Your job is two-fold: first, to solve the puzzles to ensure that they have solutions, and are unable to solve the problem on their own. This evening, one of your colleagues claimed that the following problem was unsolvable. He claims that there is no way to determine which bags the heavy and light coins are in with just one weighing and that it is also impossible to cover the checkerboard in the manner described. His reason for believing that the two parts of the puzzle are unsolvable is simply that he "couldn't find a solution." By Friday morning, the editors need you to solve this problem (if possible) and to either prove that it can't be solved or solve ir and present a written and oral solution. You have five bags of 21 coins each. Every coin should weigh 10 grams, but only the coins in three bags have the correct weight. Those of one bag weigh 9 grams each and those of another weigh 11 grams each. You need to use some of the 9 grams coins and some of the 11 gram coins to cover a 3 x 7 checkerboard in such a way that each block has exactly one coin on it, and so that the checkerboard does not contain a rectangle* whose four corners have the same type of coin. You may make use of a weighing machine whose hand shows the exact weight on a dial, but may use it only once. *Here, a rectangle is defined as consisting of more than one row and more than one column

  • Answer:

    The above answer by yungkitlee is very good and deserves to be chosen as Best Answer. To add a little to it: In fact, there are over 5000 different combinations of selections of coins from the bags that work, by producing 20 different total weights. Let a, b, c, d, e = coins chosen from the bags. Total weight = 10 * (a + b + c + d + e) + number 11g coins - number of 9g coins . So as long as every pairwise difference among a-e is different, we're good. 1 2 4 8 16 (good old powers of 2 is one) Differences are: 1,3,7,15 2,6,14 4,12 8 (These can be positive or negative, depending on which is the 9's and which is the 11's). The smallest set is 0 1 3 7 12 Differences are 1,3,7,12 2,6,11 4,9 5 Among the largest set is 10 13 14 19 21 with differences 3,4,9,11 1,6,8 5,7 2 and 9 14 18 20 21 with differences 5,9,11,12 4,6,7 2,3 1 His explanation of the second part is basically correct, but could use a little improvement on the wording. Imagine the checkboard as 7 across and 3 down: x x x x x x x x x x x x x x x x x x x x x Each column will have one of the 8 3-way AB combinations. Repetition of any of them creates a 4C rectangle, so all 7 must be different, and we can only exclude one of the 8. But each of them creates a 4C rectangle with 2 (or more) others: AAA with AAB and ABA AAB with AAA and ABB ABA with AAA and ABB BAA with AAA and BAB (and similarly for BBB, BBA, BAB, and ABB). So no matter which one is excluded, there remains a pair "in conflict" which creates a 4C rectangle.

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That is a very interesting and fun company you work for. Is your company interested in hiring by any chance? I am putting my two cents worth and I agree partly with your colleague. I solved part 1 of your question by using different amount of coins to each bag and weighing them getting the exact weight hence deducing which bag has the 11 gr and 9 gr coins respectively. Take, 1 amount of coin from the first bag, 5 amount of coins from the second, 10 amount from the third, 20 amount from the fourth and 21 amount from the 5th. You would need to mark the coins somehow from each bag so you will know which coin is from which bag. Weigh the total of 57 coins and you will get a weight. That weight will let you know the which coins are 9 gr and 11gr. For example if you get a weight of 566 gr then the first bag will have 11 gr coins and 2nd bag will have 9 gr coins. If you get a weight of 580 gr then the 3rd bag will have 9 gr coins and 4th bag will have 11 gr coins. I used an Excel sheet to produce the list. (The trick is that the number of coins from each bags have to be selected in such a s way such that the total weight for each combination is different from the rest. I am sure there are other combinations but the combination 1 coin from bag 1, 2 coins from bag 2, 3 coins from bag 3, 4 coins from bag 4 and 5 coins from bag 5 does not work). The second part of your question I agree with your colleague. There is no solution available. Let assume the two types of coins are A and B respectively. Hence to place them in groups of 3 you will get 8 combinations namely AAA BBB AAB ABA ABB BAA BAB BBA AAA and BBB are not usable as they will form a 4C rectangle (4C rectangle will mean that 4 corners of the rectangle are of the same coins) with any of the other combinations. For example AAA will form a 4C rectangle with AAB, BAA and ABA. BBB will form a 4C rectangle with ABB, BBA and BAB. Obviously no combination is to be used more than once as it will form a 4C rectangle too. That leaves you with 6 combinations and that wont be enough to fill up a 3x7 checkerboard i.e. 6 rows of 3 coins. This is one mighty interesting question. Do feel free to email me if you have any more puzzles like that. I am available for email via my profile on answers.yahoo.com

That is a very interesting and fun company you work for. Is your company interested in hiring by any chance? I am putting my two cents worth and I agree partly with your colleague. I solved part 1 of your question by using different amount of coins to each bag and weighing them getting the exact weight hence deducing which bag has the 11 gr and 9 gr coins respectively. Take, 1 amount of coin from the first bag, 5 amount of coins from the second, 10 amount from the third, 20 amount from the fourth and 21 amount from the 5th. You would need to mark the coins somehow from each bag so you will know which coin is from which bag. Weigh the total of 57 coins and you will get a weight. That weight will let you know the which coins are 9 gr and 11gr. For example if you get a weight of 566 gr then the first bag will have 11 gr coins and 2nd bag will have 9 gr coins. If you get a weight of 580 gr then the 3rd bag will have 9 gr coins and 4th bag will have 11 gr coins. I used an Excel sheet to produce the list. (The trick is that the number of coins from each bags have to be selected in such a s way such that the total weight for each combination is different from the rest. I am sure there are other combinations but the combination 1 coin from bag 1, 2 coins from bag 2, 3 coins from bag 3, 4 coins from bag 4 and 5 coins from bag 5 does not work). The second part of your question I agree with your colleague. There is no solution available. Let assume the two types of coins are A and B respectively. Hence to place them in groups of 3 you will get 8 combinations namely AAA BBB AAB ABA ABB BAA BAB BBA AAA and BBB are not usable as they will form a 4C rectangle (4C rectangle will mean that 4 corners of the rectangle are of the same coins) with any of the other combinations. For example AAA will form a 4C rectangle with AAB, BAA and ABA. BBB will form a 4C rectangle with ABB, BBA and BAB. Obviously no combination is to be used more than once as it will form a 4C rectangle too. That leaves you with 6 combinations and that wont be enough to fill up a 3x7 checkerboard i.e. 6 rows of 3 coins. This is one mighty interesting question. Do feel free to email me if you have any more puzzles like that. I am available for email via my profile on answers.yahoo.com

yungkitlee

The above answer by yungkitlee is very good and deserves to be chosen as Best Answer. To add a little to it: In fact, there are over 5000 different combinations of selections of coins from the bags that work, by producing 20 different total weights. Let a, b, c, d, e = coins chosen from the bags. Total weight = 10 * (a + b + c + d + e) + number 11g coins - number of 9g coins . So as long as every pairwise difference among a-e is different, we're good. 1 2 4 8 16 (good old powers of 2 is one) Differences are: 1,3,7,15 2,6,14 4,12 8 (These can be positive or negative, depending on which is the 9's and which is the 11's). The smallest set is 0 1 3 7 12 Differences are 1,3,7,12 2,6,11 4,9 5 Among the largest set is 10 13 14 19 21 with differences 3,4,9,11 1,6,8 5,7 2 and 9 14 18 20 21 with differences 5,9,11,12 4,6,7 2,3 1 His explanation of the second part is basically correct, but could use a little improvement on the wording. Imagine the checkboard as 7 across and 3 down: x x x x x x x x x x x x x x x x x x x x x Each column will have one of the 8 3-way AB combinations. Repetition of any of them creates a 4C rectangle, so all 7 must be different, and we can only exclude one of the 8. But each of them creates a 4C rectangle with 2 (or more) others: AAA with AAB and ABA AAB with AAA and ABB ABA with AAA and ABB BAA with AAA and BAB (and similarly for BBB, BBA, BAB, and ABB). So no matter which one is excluded, there remains a pair "in conflict" which creates a 4C rectangle.

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