Find the counterfeit coin!
You are given 9 coins which look identical, but one is lighter than the rest. Using the balance scale above, what is the minimum number of weighings you need to guarantee that you can find the counterfeit?
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11 solutions
but, What if I weigh 4 coins and the other 4 coins together and noticed that both the coins weigh the same? Isn't that 1 reading?
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what if the lighter one is in the group of 4's
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but the question says, "what is the minimum number of weighings" So, it could be 1. Therefore minimum weighing is 1.
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@Mayank Kunwar – "minimum number of weighings you need to GUARANTEE." You are not guaranteed to know from 1 draw, because the coin COULD be in any particular spot.
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@Luther Lessor – If they are all identical except one then with blind luck just by weighing one coin on each side of the scale you could find the lighter coin on the first try
@Mayank Kunwar
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Divide in the group of 3 coins.
Case 1: Measure weight of two groups of coins, if same then odd weight coin is in 3rd group.Again similar way divide the 3rd group in to the group of 1 coin each and measure weight of first two new groups. if this is same means 3rd coin is odd otherwise heavy weight coins will be odd.
Case 2: If weight of two group of 3-3 coins not same, means odd coin is present in one of the two group who will more heavy.Then select that group and again divide it into 3 parts of 1 coin each and measure weight of first two parts as above. Thus, You will find the minimum possible number of weighting = 2 to find the odd coin.
@Mayank Kunwar – That''s what I came up with as well
But if they don't weigh the same then you will have to keep going, your way doesn't guarantee finding it in one reading
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It's based on luck, yes. But if they do weight the same you actually CAN garantee, with 100% certainty that the left over coin is the lighter one.
One reading. But with 4 coins, we will need 3 times weighing at least.
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Yes,but here we need minimum no. of weighting. so, we will prefer 3 groups (3 coins each group) and thus we can find the odd weight coins in just 2 times of weighting.
That is based on the chance that you don't pick the lighter coin, and its the only one left behind.
But if you did pick it and put it on the scale, the scale will be tilted and you would have to measure again to pinpoint the lighter coin.
So, the answer on 1 measurement in based purely on chance and therefore cannot be a definite answer.
It is. And if they are both same. The remaining coin is light. I think the answer should be one. For all odd integers
agreed, that is what I came up with
You are correct. That method essentially has two weightings at one time. This is theoretically possible, which would be accurate given the question provided. I see a lot of similar questions poorly worded. For the answers they are seeking they need to say something about equal division or all coins being weighted at the same time.
I got the same solution as that
No, it says "to guarantee", you're counting on luck.
yes it is and that should have been the answer
Now I understand! It means to GUARANTEE that I will get the coin. Thank you people.
Jingyang Tan is correct.
it seems to be a good idea
cant we weigh them in groups of 4 so that if ever the lighter coin wasn't placed on the first weighing the balance scale would be balanced therefore knowing that the one coin that wasnt part is the lighter one resulting in just one wheighing.
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I agree with Mervin. Two groups of four coins, and you hold one coin back. If the two groups balance the scale, then the coin you held back is the lighter one. Yes, it's only around 11% that you'd grab the lighter one by chance, but the previously-mentioned 3-group solutions are wrong.
You need a minimum of 1 because you have to start somewhere. 1 is the minimum amount of attempts next to none but it's no attempt if you don't try.
That was my reasoning too, well explained!
You're first weighing is good but in the second weighing how can you know which coin is counterfeit? You assume it is lighter and that is not stated in the question. You would have to weigh the third coin in order to deduce which coin had a different weight. If not please explain.
Here's the solution of 2. Split the coins into 3 groups I will call them group 123 : 456 : 789. Next I will weigh 123 and 456 in the scale. One of 3 scenarios is going to happen: 1. 123 is lighter. 2. 456 is lighter or 3 they weigh the same, making 789 lighter. Next weigh guarantees the solution: take the lightest group discovered in weigh one and weigh any two coins from that group. If he scale is tilted then the lighter coin is present on the scale. If the scale is equal then the lighter coin is not on the scale.
My answer is 1, just like Jingyang Tan. If you are lucky to split the coins into 4 and and found out that the scale is balanced, then you have the fake coin in your hand. And no further balancing is necessary! THE CORRECT ANSWER IS !.
If you're very lucky to chose the counterfeit coin and a non-counterfeit coin to balance. Just one coin in each side of the balance, because you're already sure that 8 of the coins weigh the same and the counterfeit coin weighs lighter than any other. So luckily weighing one coin with a counterfeit coin determines the counterfeit coin already even at a single try. The counterfeit coin would weigh lighter than any other of the 8 coins. Therefore, it is possible to know even at a single weighing.
How bout taking 4 coins for each side
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That is a slower method. We're looking for the fastest (least weighings) method.
Two groups of four with the counterfeit coin being left out. If the two groups of four are balanced we know the remaining coin is counterfeit. Thus being done with only one weighing..
SERIOUSLY?!? AGAIN, If the fake one is lighter than the real one, but we aren't told that the other 8 weigh the same, it's not possible to determine the answer in TWO weighs!!!!
I think the "minimum" is one. Put 4 coins on each side of the scale to begin with. If it balances out at the first try, then one will know the fake coin. If it doesn't balance out, remove one coin at a time from each side until the scale balances out. I thought it was supposed to be a logic puzzle....
The ans should be 3 if divide in 2groups of 4 then taking odds against us one group is lighter.. Then we weigh two times in a group of 2 and 1 resp. your ans is wrong... If we split in group of 3 we get 3groups if we consider for guarantee we have to assume the first 2 groups balance then balance once again with 1on each side and one left behind if they get balanced then the left behide one is lighter or the ligter weighing...
Why not 1 - put four coins on each side. If they balance then you know the one off the scale is the counterfeit. That makes one the (unlikely) minimum. perhaps the phrasing of the question needs refinement?
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The question says u need to "guarantee" here u r relying on luck that counterfeit coin won't be in any of the group of 4
The minimum number of weighing should be 1 (it does however require some luck). Choose two coins and place them on each side. If one of them is the counterfeit you will know after the first weighing.
You are of course not guaranteed to pick the counterfeit the first time, and will have to try again, but the question asks for the MINIMUM number.
If we don't know whether coin is light or heavy then formula would be.. log3(2n+1)<=k , k is no of trials and n is number of balls.
Yes, I believe so that the answer is 2 but I used mind instead. I also believe maximum 2 weightings will be required as if you split the coins in groups of 3, in one weighing you will find out which has the fake coin. Then by one more weighing you can easily tell which is the fake one.
By placing 4 coins on each side and leaving one out. If the scale balances then the odd coin out is the lightest one.
What about, if there are 12 coins?. This was the question asked to me in a test.
Divide in the group of 3 coins.
Case 1: Measure weight of two groups of coins, if same then odd weight coin is in 3rd group.Again similar way divide the 3rd group in to the group of 1 coin each and measure weight of first two new groups. if this is same means 3rd coin is odd otherwise heavy weight coins will be odd.
Case 2: If weight of two group of 3-3 coins not same, means odd coin is present in one of the two group who will more heavy.Then select that group and again divide it into 3 parts of 1 coin each and measure weight of first two parts as above.
Thus, You will find the minimum possible number of weighting = 2 to find the odd coin.
You have asked about the minimum number of weighings. If you put 4 coins each on both pans and they weigh equal, in that case you will know that the 9th coin is lighter. thus it can be determined in minimum one weighing, so the correct answer should certainly is ONE and all above answers are wrong.
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Actually, yes. I agree! It did stipulate 'MINIMUM'....although unlikely to get it right as a one-off go.....ONE is the MINIMUM.
These questions with he word "minimum" are confusing. The previous daily puzzle had an answer of 2, but the "correct" was 3. In this one, it again asks for the minimum, which is one. If you split the coins 4 and 4 with one left over, and the scale balances, you are guaranteed the leftover coin is the light one. Is it likely? No, but it is the minimum.
2 times minimum. Split them as three of three groups. Place three on one side of scale and three on other side of the scale, three not on the scale. If one side the scale has less weight you can see the lighter one is in that group . If the scale is balancing you can understand the lighter one is in the group of three not on the scale. Take the group of three with the lighter coin and put one coin on one side side, one coin on the other side, one coin off the scale. From this we can find lighter one. If they have same weight, Lighter one is not on the scale.
Firstly, we must divide the coins into three different groups, which we will call x , y and z . We can then take two of the three groups. Let's say that we take x and y , and we put z to the side. Since only one of the coins has a greater weight, we will know that if one of the sides on the scale drops lower than the other, then one of these groups contains the coin. If it balances, we will know that group z contains the heavier coin. After doing this, we will then know which group has the coins. Let's say that we will rename the group of coins from x , y or z to the name, group a . We all then take group a and split it into three different groups of one coin. We will take two of the three coins and put them on the scale, leaving the other one to the side. Again, the same principle for the groups applies to this situation as well. If one of the sides on the scale lowers down, we will know that the lowered side has the coin. If the sides are balanced, then the coin is the one that was set aside.
This method is the most efficient because the coins are initially split into equal, quite small groups, where we are not only utilizing the scale, but the information that the scale tells us about the group that is not on the scale. If we try to do this with any other combination of coins, it is always higher because of the amount of coins in each of the groups or the ways which they can be divided.
Its the minimum number required. If I put 4 coins oneach side and it is balanced, then the coin not included is the light one. Got to be lucky , but that is the MINIMUM. One weighing.
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It is the minimum number of weighings that GUARANTEES finding the counterfeit coin. Your method work work in some cases, but in most situations, it wouldn't guarantee finding the coin.
How do i know that the lighter one is counterfeit, what if the other 8 are the counterfeit ones?
The question of illegal or counterfeit is secondary to the problem posed by the question.
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Not at all! A very 'thinking-outside-the-box' response! 😊
Very realistic but you need to READ THE PROBLEM. This is not the real world(otherwise you may be asking why the forger is telling you he's cheating you).
First, put the coins into 3 piles of 3 coins. Next, Put 2 piles of the coins on the scale. 2 scenarios will happen: 1. If the balance is equal, then the 3 piles has the lightweight coin. 2. If the balance is different, then one with the shorter weight has the lightweight coin. Now you have to put 2 coins on the scale. again, 2 scenarios will happen. 1. If the balance is equal, then the 3rd coin is the lightweight. 2. If the balance is different, then the one with less weight is the lightweight.
Split the coins into 3 equal piles. Weigh two piles. If one side is lighter, that side has the light coin. If balanced, the pile you didn't weigh has the light coin. Now, weigh two coins in the correct group against each other. If one is lighter, that's the light coin. If the second weighing gives you a balance, the third coin in the correct pile is lighter than the other coins. Thus, the answer is 2 .
First of all I agree with one of the subscribers about 'lighter' and 'counterfeit'
The wording should be modified. Itshould either say 'counterfeit is lighter' or should have asked '...find the lighter one' at the end.
Secondly the people giving the answer as 1 have concentrated on "minimum" and totally missed "guaranteed". If their line of argument is to be accepted then for a zillion number of coins only one weighing is needed. All you need to do is pick any two coins and place one in each pan. Viola! With a one-in-zillion (or two in a zillion to be more accurate) chance one is lighter and so that is the coin we are looking for. So yes you may find it in one weighing but does that guarantee it?
divide into three A=3,B=3,c=3 step 1: weight A and B if A==B then the lighter coin is in C IF A is not equal to B then lighter one contain the defect coin from this we can get one of the group contain lighter oe Step 2: divide into three sections again it will D=1;E=1 F=1 weight D and E if D==E lighter is F else which once is lighter than the final coin
first consider coins as no.s then weigh first 1 and 2 .think that 2 is light then it will be the answer as it is telling one is lighter than the rest.notice:if my answer is wrong please correct me
It's in the worst case scenario
Split the coins into 3 groups of 3 (call these groups A, B, and C).
Weighing #1: First, put A and B on the two sides of the scale. If one of the sides of the scale is higher (lighter), then this group of 3 coins contains the counterfeit. If the scale is balanced, then group C contains the counterfeit.
Weighing #2: Now, we have narrowed it down to 3 coins with one counterfeit. Place one coin on each side of the scale. If one of the sides of the scale is higher (lighter), then that is the counterfeit. If the scale is balanced, then the third coin (not on the scale) must be the counterfeit.
Thus, we can guarantee finding the counterfeit in just 2 weighings!
Note : Some comments have suggested that 1 weighing would work, because you could put 4 and 4 and if they happen to be equal, then the unweighed coin is the counterfeit. However, the problem asks for the "minimum number of weighings you need to guarantee that you can find the counterfeit". The strategy of 4 and 4 would usually not work, so you cannot guarantee that you can find the counterfeit in one weighing.