Posted 18 Jun 2007 · Report post You've got 27 coin, each of them is 10 g, except for 1. The 1 different coins has a masses different. It could be 9 g, or 11 g (heavier, or lighter by 1 g). The question is, in how many minimum ways, you can determine the different coins, and is it heavier, or lighter? I have no answer to this question. Do you? 0 Share this post Link to post Share on other sites

Posted 18 Jun 2007 · Report post Assuming you have a simple scale, my answer is 9. First divide the coins into 3 piles of 9. Measure 2 piles at a time until you find the pair that is equal in mass. This can take up to 3 tries, but if you were lucky you could get it on the first try. Put the piles of coins with equal mass to the side. Repeat the process this time with 3 piles of 3 coins, maximum 3 measurements. Repeat the process this time with the 3 remaining coins, find the two that are equal, the remaining coin has a different mass. If you are a very luck person you coulp potentially only have to make 4 meaurements, 3 to rule out equal pairs and 1 to compare the final coin to the rest. 9 is the most you would have to do the efficient way, and at any point you can observe is the unequal pile has more or less mass. 0 Share this post Link to post Share on other sites

Posted 16 Jan 2008 · Report post I guess the question is lack of some information such as, by using a scale with weight of 1g, 2g, 5g, and 10g, isn't it? 0 Share this post Link to post Share on other sites

Posted 16 Jan 2008 · Report post I guess the question is lack of some information such as, by using a scale with weight of 1g, 2g, 5g, and 10g, isn't it? It's a balance scale that compares what's in the two pans. You can get the answer by just comparing groups of coins. 0 Share this post Link to post Share on other sites

Posted 16 Jan 2008 · Report post Duplicate post. Looking around the forum for an eraser ... <_< 0 Share this post Link to post Share on other sites

Posted 18 Jan 2008 · Report post 1-9 vs 10-18 -if it balances, the odd one is in 19-27 -if it doesn't balance: 0 Share this post Link to post Share on other sites

Posted 18 Jan 2008 · Report post 1-9 vs 10-18 -if it balances, the odd one is in 19-27 -if it doesn't balance: 0 Share this post Link to post Share on other sites

Posted 18 Jan 2008 · Report post sorry - accidentally hit submit - here's my answer... this can be done in 5 weighings: weigh coins [1-9] vs [10-18] (ie coins 1 thru 9 on one side, coins 10 thru 18 on the other) if it balances, the odd one is in 19-27 if it doesn't balance, weigh [1-9] vs [19-27] -if it balances, the odd one is in 10-18 -if it doesn't balance, the odd one is in 1-9 after 2 weighings we know which group of 9 has the odd one and weather the odd coin is heavier or lighter than the others. For the remainder of the explanation, I'll assume it's in group 1-9, but the logic will work for either of the other groups. weigh [1-5] vs [6-9 and 27] (ie coins 1 thru 5 on one side, 6 thru 9 and 27 on the other. For this example, know #27 is ok, if the odd one is in the 19-27 group, we'd pick one of the good coins from another group to include instead) after the 2nd weighing we learned whether to look for heavier or lighter group, so pick either 1-5 or 6-9 as having the odd coin based on heavy vs light whichever group the odd one is in, weigh 2 coins vs 2 coins for example for 1-5, weigh [1-2] vs [3-4] - if they balance its #5, if not pick the correct side based on heavier or lighter learned in weigh #2 and weigh them against each other alternatively, if it's in 6-9, weigh [6-7] vs [8-9], pick the heavier or lighter pair based on weigh #2 and weigh them against each other. This process should work regardless of which one it is, and whether it's lighter or heavier. 0 Share this post Link to post Share on other sites

Posted 18 Jan 2008 · Report post I think I've got it in 4 measurements: Separate the coins into 3 stacks of 9 (A, B, C). Weigh stack A against B and then A against C. Take the stack with the different weight (note lighter or heavier) and break it into 3 stacks of 3 (D, E, F). Weigh stack D against E. If D and E are equal, then F is the odd stack. If D and E are not equal, the lighter or heavier (based on the A, B, C comparison) is the odd stack. You now have three coins (G, H, I). Weigh G and H. If G equals H, then I is the odd and is lighter or heavier (based on the A, B, C comparison). If G and H are not equal, then the lighter or heavier (based on the A, B, C comparison) is the odd coin. 0 Share this post Link to post Share on other sites

Posted 18 Jan 2008 · Report post Jkyle1980's solution is correct, and much less verbose than what follows. You can distinguish among 3**N cases in N weighings. There are 54 possible cases in this puzzle [one of 27 coins is heavy or light]. So three weighings [27 cases] won't do it, but four [81 cases] can. Use Jkyle1980's groups of nine coins: A, B, C. [1] A = B: C has a heavy or light coin [18 cases] [2] A < B: A has a light coin or B has a heavy coin [18 cases] [3] A > B: A has a heavy coin or B has a light coin [18 cases] [1] Split C into groups of three coins: C1, C2, C3 and weigh C1 vs. C2. C1 = C2: C3 has a heavy or light coin [6 cases] Name the C3 coins c3a, c3b, c3c. c3a = c3b: if c3a </> c3c, then c3c is the heavy / light coin. c3a < c3b: if c3a </= c3c, then c3a is light / c3b is heavy. c3a > c3b: if c3a =/> c3c, then c3b is light / c3a is heavy. C1 < C2: C1 has a light coin or C2 has a heavy coin [6 cases] C1 = C3: [C2>C1=C3] C2 has a heavy coin. Procedure 1: Name the coins c2a, c2b, c2c. c2a = c2b: c2c is heavy c2a < c2b: c2b is heavy c2a > c2b: c2a is heavy C1 < C3: [C2>C1<C3] C1 has a light coin. Apply Procedure 1 to determine the light coin. C1 > C3: [C3>C1>C2] Impossible. C1 > C2: C1 has a heavy coin or C2 has a light coin [6 cases] C1 = C3: [C1=C3>C2] C2 has a light coin. Apply Procedure 1 to determine the light coin. C1 < C3: [C2<C1<C3] C1 Impossible. C1 > C3: [C2<C1>C3] C1 has a heavy coin. Apply Procedure 1 to determine the heavy coin. [2] Weigh A vs. C A = C: [A=C<B] B has a heavy coin [9 cases] Procedure 2: Split the B coins into groups of three: B1, B2, B3 B1 = B2: B3 has a heavy coin. Apply procedure 1 to determine which. B1 < B2: B2 has a heavy coin. Apply procedure 1 to determine which. B1 > B2: B1 has a heavy coin. Apply procedure 1 to determine which. A < C: [b>A<C] A has a light coin [9 cases] Apply Procedure 2 to determine which. A > C: [b>A>C] Impossible [3] Weigh A vs. C A = C: [A=C>B] B has a heavy coin [9 cases] Apply Procedure 2 to determine which. A < C: [b<A<C] Impossible A > C: [b<A>C] A has a heavy coin [9 cases] Apply Procedure 2 to determine which. 0 Share this post Link to post Share on other sites

Posted 18 Jan 2008 · Report post Jkyle1980's solution is correct, and much less verbose than what follows. You can distinguish among 3**N cases in N weighings. There are 54 possible cases in this puzzle [one of 27 coins is heavy or light]. So three weighings [27 cases] won't do it, but four [81 cases] can. Use Jkyle1980's groups of nine coins: A, B, C. [1] A = B: C has a heavy or light coin [18 cases] [2] A < B: A has a light coin or B has a heavy coin [18 cases] [3] A > B: A has a heavy coin or B has a light coin [18 cases] [1] Split C into groups of three coins: C1, C2, C3 and weigh C1 vs. C2. C1 = C2: C3 has a heavy or light coin [6 cases] Name the C3 coins c3a, c3b, c3c. c3a = c3b: if c3a </> c3c, then c3c is the heavy / light coin. c3a < c3b: if c3a </= c3c, then c3a is light / c3b is heavy. c3a > c3b: if c3a =/> c3c, then c3b is light / c3a is heavy. C1 < C2: C1 has a light coin or C2 has a heavy coin [6 cases] C1 = C3: [C2>C1=C3] C2 has a heavy coin. Procedure 1: Name the coins c2a, c2b, c2c. c2a = c2b: c2c is heavy c2a < c2b: c2b is heavy c2a > c2b: c2a is heavy C1 < C3: [C2>C1<C3] C1 has a light coin. Apply Procedure 1 to determine the light coin. C1 > C3: [C3>C1>C2] Impossible. C1 > C2: C1 has a heavy coin or C2 has a light coin [6 cases] C1 = C3: [C1=C3>C2] C2 has a light coin. Apply Procedure 1 to determine the light coin. C1 < C3: [C2<C1<C3] C1 Impossible. C1 > C3: [C2<C1>C3] C1 has a heavy coin. Apply Procedure 1 to determine the heavy coin. [2] Weigh A vs. C A = C: [A=C<B] B has a heavy coin [9 cases] Procedure 2: Split the B coins into groups of three: B1, B2, B3 B1 = B2: B3 has a heavy coin. Apply procedure 1 to determine which. B1 < B2: B2 has a heavy coin. Apply procedure 1 to determine which. B1 > B2: B1 has a heavy coin. Apply procedure 1 to determine which. A < C: [b>A<C] A has a light coin [9 cases] Apply Procedure 2 to determine which. A > C: [b>A>C] Impossible [3] Weigh A vs. C A = C: [A=C>B] B has a heavy coin [9 cases] Apply Procedure 2 to determine which. A < C: [b<A<C] Impossible A > C: [b<A>C] A has a heavy coin [9 cases] Apply Procedure 2 to determine which. So, like I said, 4 right? 0 Share this post Link to post Share on other sites

Posted 18 Jan 2008 · Report post I think I've got it in 4 measurements: Separate the coins into 3 stacks of 9 (A, B, C). Weigh stack A against B and then A against C. Take the stack with the different weight (note lighter or heavier) and break it into 3 stacks of 3 (D, E, F). Weigh stack D against E. If D and E are equal, then F is the odd stack. If D and E are not equal, the lighter or heavier (based on the A, B, C comparison) is the odd stack. You now have three coins (G, H, I). Weigh G and H. If G equals H, then I is the odd and is lighter or heavier (based on the A, B, C comparison). If G and H are not equal, then the lighter or heavier (based on the A, B, C comparison) is the odd coin. silly me - I was so excited to get a better answer than 9 weighings , that I missed the simplicity of just 4. Good job jkyle1980... 0 Share this post Link to post Share on other sites

Posted 1 Feb 2008 · Report post You've got 27 coin, each of them is 10 g, except for 1. The 1 different coins has a masses different. It could be 9 g, or 11 g (heavier, or lighter by 1 g). The question is, in how many minimum ways, you can determine the different coins, and is it heavier, or lighter? I have no answer to this question. Do you? In fact the answer is 2!!!! You need a minimun of 2 weighs to find the bad coin IF YOU ARE LUCKY! 1- You take 2 coins, you weight them and you get an unbalanced situation. 2- You take one of them and you compare to an outside coin and you still find an unbalanced situstion ==> The bad coin is X and its 9g or 11g So the minimum ways needed are 2. Since we already know this type of problems, it's easy to assume what was meant by the OP. However I'm sure that what was meant originally was: What is the minimum number of weighs needed to guarantee finding the "Bad" coin and its weight are... 4 Well, if we knew the real weight of the "bad" coin (9 or 11) we would be able to do it in 3 weighings for 27 coins (discussed previously on this forum) But in our case here, all we need is ONE extra weighing when we get to an unbalanced situation in any of our weighings (Let's say with groups A & B): We take one of the two unbalanced groups (A) and compare it to a group of the same number of coins from the rest of the coins (ALL GOOD COINS). This will tell us which group amongst A & B contained the bad coin AND the real weight of that coin (heavier or lighter). Then we proceed with the rest of the steps... 0 Share this post Link to post Share on other sites

Posted 5 Mar 2008 · Report post Folks, This one took a little testing before I came up with the correct answer . . . FOUR tests. Some of the flaws I've seen are getting to the final nine in the first weighing. If Group A is the same as Group B, then Group C has the heavy OR light coin. If B was heavier (just to pick one case) however, we need a second test. Test two would then compare Group B against Group C. If these are the same, then A contains a Light Coin; if B & C are equal, then B contains a heavy coin. The next two rounds would play out differently depending upon what we learned in Test 1, but perhaps it is easier to show my decision making "system." ### Let's assume you know that Coin #6 is a light coin, but I know nothing. Test 1 1-9 vs. 10-18, Group A is lighter. Test 2 10-18 vs. 19-27, these are the same, therefore I now know two facts. The odd coin is light, and the odd coin is in Group A. Remember if B had been heavier in test two, the odd coin would have been a heavy one in Group B -- You knew that the odd coin was light and only now do I know it as well. Test 3 1-3 vs. 4-6. Group 2 is Lighter. Now since we know that the coin had to be light, we are down to a final three. If Group 1 had been lighter then 1-3 would be the final group, if they had been the same then Group 3 would be the final group. Test 4 4 vs. 5. They are the same, so I now know what you knew all along, Coin #6 is light. Could you get the answer in fewer tests? A qualified NO. Here is another variation. Assume coin 27 is heavy . . . Test 1 1-9 vs. 10-18, weigh the same so Group C has the odd coin (lighter? heavier? we don't yet know). Test 2 19-21 vs. 22-24, weigh the same so Group 3 has the odd coin (lighter? heavier? we don't yet know). Test 3 25 vs. 26, weigh the same so coin #27 is the odd coin (lighter? heavier? we STILL don't know). Test 4 26 vs. 27, Coin 27 is the heavy one. Still Four tests. Basically you need three tests to definitively determine the odd coin, and a fourth test to determine whether the odd coin is heavy or light. The order in which you learn these facts is irrelevant. Note in the first example we learned two of the critical facts after Test 2, and then learned one more fact in each Test 3 and Test 4. In the second example we learned one fact in each of the four tests. The order didn't matter. So the reason I say a "Qualified No," is that a previous respondent stated that he could be lucky with just two Tests. This is correct. Assume again that coin 27 is the heaviest, and he is lucky enough to choose the following tests. Test 1 26 vs. 27, Coin 27 is heavIER (either 27 could be the heaviest coin, or 26 could be the lightest) Test 2 25 vs. 27, Coin 27 is heavier, so he can then conclude that Coin #27 is the HEAVIEST coin. But if he plays the wild guessing game, and I use my "system," we can easily conclude that analytical thinking will win out over time. I will always KNOW the relative weight of the odd coin in FOUR tests. He may GUESS it in TWO, but if he is really unlucky he may require as many as FOURTEEN tests (to prove the worse case, again 27 is heavy): Test 1: 1 vs. 2 Same Test 2: 3 vs. 4 Same Test 3: 5 vs. 6 Same Test 4: 7 vs. 8 Same Test 5: 9 vs. 10 Same Test 6: 11 vs. 12 Same Test 7: 13 vs. 14 Same Test 8: 15 vs. 16 Same Test 9: 17 vs. 18 Same Test 10: 19 vs. 20 Same Test 11: 21 vs. 22 Same Test 12: 23 vs. 24 Same Test 13: 25 vs. 26 Same Test 14: 1 vs. 27, 27 is heavier, and therefore the heaviest. His is not exactly a winning system! 0 Share this post Link to post Share on other sites

Posted 10 Mar 2008 · Report post I think I've got it in 4 measurements: Separate the coins into 3 stacks of 9 (A, B, C). Weigh stack A against B and then A against C. Take the stack with the different weight (note lighter or heavier) and break it into 3 stacks of 3 (D, E, F). Weigh stack D against E. If D and E are equal, then F is the odd stack. If D and E are not equal, the lighter or heavier (based on the A, B, C comparison) is the odd stack. You now have three coins (G, H, I). Weigh G and H. If G equals H, then I is the odd and is lighter or heavier (based on the A, B, C comparison). If G and H are not equal, then the lighter or heavier (based on the A, B, C comparison) is the odd coin. Nice one ... 0 Share this post Link to post Share on other sites

Posted 15 Mar 2008 · Report post 2 tests.. Divide into 3 piles Compare 2 piles with the third? Two of the piles has equal weight, when you find them you only need to compare one of them with the last to figure out if the coin is heavier or lighter - let one pile stay on the scale. 1st try: 2 equal piles 2nd try: the scale goes up or down OR 1st try: the scale goes up or down 2nd try: the scale has to go same as 1st - Not sure if Iâ€™ve explained it properly, my English is kinda bad <_< 0 Share this post Link to post Share on other sites

Posted 31 Mar 2008 · Report post I suppose this can be done in a simpler way than discussed above! Make 2 stacks each time and put them on the balancing pan. Lets name the two sides of balancing pan as A&B. This is how it can be done- 1. Make 2 stacks of 13 coins each and put on A&B. Now if U'r lucky, both the pans would weigh same & thus the leftover coin is ur choice! BUT, If Pan B is heavy- 2. Then take the 13 coins from pan B, divide in 6 coins each & place on A&B again. IF u'r lucky then A&B would weigh equal & the left over coin is heavier one! BUT If either side of pan is heavy (lets take A this time!) 3. Take the 6 coins from A, divide into 3 coins each & place on A&B again! No luck here buddy ! One of either side of pan would be heavy (lets take B this time!) 4. Take the 3 coins, keep one coin on each side of A&B. WOW....here u get ur coin...If one side gets heavy, then the heavier side is ur coin!...& if both weigh equal, then the leftover coin is the heavy one!!! I suppose this is a much simpler solution than the ones dealing in making 3 stacks & comparing them again & again <_< . Those solutions really would be quite painstaking, confusing & require lot of Permutations & Combinations. 0 Share this post Link to post Share on other sites

Posted 31 Mar 2008 · Report post I suppose this can be done in a simpler way than discussed above! Make 2 stacks each time and put them on the balancing pan. Lets name the two sides of balancing pan as A&B. This is how it can be done- 1. Make 2 stacks of 13 coins each and put on A&B. Now if U'r lucky, both the pans would weigh same & thus the leftover coin is ur choice! BUT, If Pan B is heavy- 2. Then take the 13 coins from pan B, divide in 6 coins each & place on A&B again. IF u'r lucky then A&B would weigh equal & the left over coin is heavier one! BUT If either side of pan is heavy (lets take A this time!) 3. Take the 6 coins from A, divide into 3 coins each & place on A&B again! No luck here buddy ! One of either side of pan would be heavy (lets take B this time!) 4. Take the 3 coins, keep one coin on each side of A&B. WOW....here u get ur coin...If one side gets heavy, then the heavier side is ur coin!...& if both weigh equal, then the leftover coin is the heavy one!!! I suppose this is a much simpler solution than the ones dealing in making 3 stacks & comparing them again & again <_< . Those solutions really would be quite painstaking, confusing & require lot of Permutations & Combinations. Jus an addition is dat this is the solution if v consider the coin to be heavy one. Same can be applied if the coin is lighter one!!! 0 Share this post Link to post Share on other sites

Posted 31 Mar 2008 · Report post I suppose this can be done in a simpler way than discussed above! Make 2 stacks each time and put them on the balancing pan. Lets name the two sides of balancing pan as A&B. This is how it can be done- 1. Make 2 stacks of 13 coins each and put on A&B. Now if U'r lucky, both the pans would weigh same & thus the leftover coin is ur choice! BUT, If Pan B is heavy- 2. Then take the 13 coins from pan B, divide in 6 coins each & place on A&B again. IF u'r lucky then A&B would weigh equal & the left over coin is heavier one! BUT If either side of pan is heavy (lets take A this time!) 3. Take the 6 coins from A, divide into 3 coins each & place on A&B again! No luck here buddy ! One of either side of pan would be heavy (lets take B this time!) 4. Take the 3 coins, keep one coin on each side of A&B. WOW....here u get ur coin...If one side gets heavy, then the heavier side is ur coin!...& if both weigh equal, then the leftover coin is the heavy one!!! I suppose this is a much simpler solution than the ones dealing in making 3 stacks & comparing them again & again <_< . Those solutions really would be quite painstaking, confusing & require lot of Permutations & Combinations. the problem is that it is not clear whether the different coin is heavier or lighter 0 Share this post Link to post Share on other sites

Posted 5 Apr 2008 · Report post You've got 27 coin, each of them is 10 g, except for 1. The 1 different coins has a masses different. It could be 9 g, or 11 g (heavier, or lighter by 1 g). The question is, in how many minimum ways, you can determine the different coins, and is it heavier, or lighter? I have no answer to this question. Do you? easy. you ask the coin if it weighed itself lately and to tell you its weight 0 Share this post Link to post Share on other sites

Posted 9 May 2008 · Report post The way I think, I use the solutions alredy given. I think in 6 weighings one can find the defective one. 1. Divide as mentioned 27 to 3 equal heaps of 9. 2. Weigh any two heaps. If they are equal , then the third heap contains the defective coin. 3. If unequal change the heap on one pan with the left over heap. If they are equal the left over contains defective coin. if the position of the scale is same then the heap on the opposite pan of the scale contains defective coin. If heap count is 1 goto 6. 4. Take the heqp with defective coin and make 3 equal heaps. Repeat steps 2 and 3. Goto 5 when the heap count is 3, else goto 2. 5. Find the defective haeap of 3coins from step 4 and repeat steps 2 and 3 6. Find the defective coin. Seeing the position of the defective coin against good coin in scale pan , one can find whether it is heavy or light Total 6 weighings will be required. This logic can be extended to any number = 3 ^N. 0 Share this post Link to post Share on other sites

Posted 4 Jun 2008 · Report post I have a minimum of 1 and a maximum of 4. Begin by removing one coin, then weigh the two sets of 13. If they are equal then the removed coin is the culprit. If not, then take the set that does not weigh 130 g and repeat the procedure. Remove one, and weigh the two sets of 6. If one of the sets is not equal to 60g, then weigh each of the 3 coins in that set. For the set not weighing 30 g, remove one and weigh each of the other two, and you will find the over/underweight coin. Agree? 0 Share this post Link to post Share on other sites

Posted 14 Jun 2008 · Report post I think I can do it with a best-case of three, and a worst case of four. If you always partition the coins into three groups, and weigh two groups against each other AND if you always get a result of "equals" Then you always know that the coins you didn't weigh are flawed, and so can achieve the result in THREE steps. (Best case). For the other cases: Let us say that the coins are numbered 0-26 If weighing 0-8 vs 9-17 gives us a value of unequal we know that one of them must be flawed, but cannot tell which. So, our next weighing is 0-2,9-11 against 12-14,3-5. What we're doing is taking three coins from each pile and setting them aside, and then switching three coins from their pans. If the pans don't change their balance, the flawed coin is in the group that didn't switch, i.e. 0-2, 12-14 If the pans change balance, the flawed coin is in the group that did switch, i.e. 3-5,9-11 If the pans balance, the flawed coin is in the group that was set aside, i.e. 6-8, 15-17 Either way, in 2 steps we've eliminated all but 6 coins. From these six coins, if they were NOT the ones set aside, i.e. they are from {0-5, 9-14} renumber them as 0-2 for the ones that came from the left pan, and 3-5 for the ones that came from the right pan. weigh 0,3 against 1,4 if they tilt, the flawed coin is in 1,3, i.e. the coins that moved. if they don't change, the flawed coin is in 0,4, i.e. the coins that didn't move if they balance exactly, the flawed coin is in 2,5, i.e. the coins that weren't weighed. It will take a further weighing against an ideal specimen to find the flawed coin, taking FOUR steps If the coins WERE set aside, remember which side was lighter than the other, do the swap/remove as above, and measure whether the pans tilt, stay the same, or balance. Takes FOUR steps in all cases except for the best. If we could somehow only do three-way comparisons, we could always do it in log3 (27) = 3 steps, but I guess 4 steps in general is good. You can think of it as 3 steps to find out which coin it might be, and 1 step to figure out if it's heavier or lighter, which is not needed if we never use the coin. Anyway, happy sorting! Nvader 0 Share this post Link to post Share on other sites

Posted 15 Jun 2008 · Report post Separate the coins into 13 each and one more. Put 13 on each side of balance, if equal, the odd one is the one with a different weight. If one is more or less than 130 Gms that bundle is taken for further scrutiny. The thirteen in that each are similarly divided into 6, 6 and one to find out if one left out is the culprit. If one of the "sixes" weighs more or less than 60 gms that bundle is divided into 3 each to find out where is the odd one. Out of the three with one odd, take out one and weigh the other two and we aill definitely get the odd one. Hejnce with a minimum of weighing 4 times we can get the odd coin out. 0 Share this post Link to post Share on other sites

Posted 28 Jun 2008 (edited) · Report post You've got 27 coin, each of them is 10 g, except for 1. The 1 different coins has a masses different. It could be 9 g, or 11 g (heavier, or lighter by 1 g). The question is, in how many minimum ways, you can determine the different coins, and is it heavier, or lighter? I have no answer to this question. Do you? out of the 27 coins, make 3 piles of 9 coins each. weigh each pile against each other. one pile should be different (in other words, two of the piles should be perfectly balanced. the pile that tips the scale is set aside for the next step). then take that 9, and divide into 3 piles again of 3 coins each and do the same. lastly, take the resulting different pile with the last remaining 3 coins,and weigh them against each other. (so, if you put one coin on each side, and it's balanced you know the one you didn't put on the scale is the "different" coin. if it's unbalanced, switch one of the coins until it is.) tah-dah! right?? Edited 28 Jun 2008 by janica 0 Share this post Link to post Share on other sites