You have twelve coins. You know that one is fake. The only thing that distinguishes the fake coin from the real coins is that its weight is imperceptibly different. You have a perfectly balanced scale. The scale only tells you which side weighs more than the other side.
What is the smallest number of times you must use the scale in order to always find the fake coin?
Use only the twelve coins themselves and no others, no other weights, no cutting coins, no pencil marks on the scale. etc.
These are modern coins, so the fake coin is not necessarily lighter.
Presume the worst case scenario, and don't hope that you will pick the right coin on the first attempt.
Solution:
3.
If you knew the fake coin was lighter, then the solution would have an easy explanation. But you do not. So....
Number the coins 1 through 12.
1. Weigh coins 1,2,3,4 against coins 5,6,7,8.
1.1. If they balance, then weigh coins 9 and 10 against coins 11 and 8 (we know from the first weighing that 8 is a good coin).
1.1.1. If the second weighing also balances, we know coin 12 (the only one not yet weighed) is the counterfeit. The third weighing indicates whether it is heavy or light.
1.1.2. If (at the second weighing) coins 11 and 8 are heavier than coins 9 and 10, either 11 is heavy or 9 is light or 10 is light. Weigh 9 against 10. If they balance, 11 is heavy. If they don't balance, you know that either 9 or 10 is light, so the top coin is the fake.
1.1.3 If (at the second weighing) coins 11 and 8 are lighter than coins 9 and 10, either 11 is light or 9 is heavy or 10 is heavy. Weigh 9 against 10. If they balance, 11 is light. If they don't balance, you know that either 9 or 10 is heavy, so the bottom coin is the fake.
1.2. Now if (at first weighing) the side with coins 5,6,7,8 are heavier than the side with coins 1,2,3,4. This means that either 1,2,3,4 is light or 5,6,7,8 is heavy. Weigh 1,2, and 5 against 3,6, and 9.
1.2.1. If (when we weigh 1,2, and 5 against 3,6 and 9) they balance, it means that either 7 or 8 is heavy or 4 is light. By weighing 7 and 8 we obtain the answer, because if they balance, then 4 has to be light. If 7 and 8 do not balance, then the heavier coin is the counterfeit.
1.2.2. If (when we weigh 1,2, and 5 against 3,6 and 9) the right side is heavier, then either 6 is heavy or 1 is light or 2 is light. By weighing 1 against 2 the solution is obtained.
1.2.3. If (when we weigh 1,2, and 5 against 3, 6 and 9) the right side is lighter, then either 3 is light or 5 is heavy. By weighing 3 against a good coin the solution is easily arrived at.
1.3 If (at the first weighing) coins 1,2,3,4 are heavier than coins 5,6,7,8 then repeat the previous steps 1.2 through 1.2.3 but switch the numbers of coins 1,2,3,4 with 5,6,7,8.
Comments (77)
The problems were quite challenging. The idea of partitioning different levels is appreciable.
The last puzzle was too easy; I solved it in under 30 seconds. Please make the puzzles harder.
The robot's response to a simple question about how many fingers are being held up would require it to lie.
A hint for the last puzzle would be helpful, as I struggled with it and had to guess the answer.
I misunderstood the problem. I thought it was implying that you would be moving upstream at three miles per hour, but it actually explains why you should ignore the speed of the stream.
I guessed the right answer, but it seems incorrect because the answer is given before rolling. Am I misunderstanding the puzzle?
In 'The Stark Raving Mad King', if there are 10 wise men and one can see all the caps in front of him, he can deduce his own cap color. This reasoning continues for the others, leading to a logical conclusion.
The answer to the cup handle question is indeed outside, but it might be worth discussing why that is the case.
I cannot understand the answer being 3. If we cannot distinguish if the fake coin is lighter or heavier, would this not add an extra use of the scale? Can you please let me know the solution?
Your solution works when asking the truthful robot, but fails with the liar robot. Please clarify my understanding.
I have an alternate solution that is easier: Separate coins into piles A, B, C, and D. Measure A against B and B against C. If A and B balance but B and C don't, the coin is in C. If A and B are off but B and C balance, the coin is in A. If neither measurement balances, the coin is in B. If both balance, the coin is in D. You now have it narrowed down to 3 coins.
The alternate solution for The Fake Coin is much easier: Separate coins into piles A, B, C, and D. Measure A against B and B against C to narrow down the fake coin.
Difficult #7 Solution seems to be incorrect. When creating 2 sets of ten coins, you are choosing a subset of the 100 coins without knowing heads/tails. A counter example is if both sets of 10 coins were all heads, you'd then end up with a set of 10 tails and a set of 10 heads.
The burn time of the strings can vary significantly based on their random burn rates. If one string takes 5 minutes to burn to a point and 55 minutes to burn to the other end, lighting both ends simultaneously could lead to the flames meeting much sooner than expected.
The solution to the 12 coins problem incorrectly assumes the heavier coin is the counterfeit without confirming that. Additionally, the first 100 coins problem does not specify the number of pirates.
The second solution doesn't explain how a circuit could be made with wire 1 when the technician leaves the battery hook up at one end and uses the bulb at the other. Regarding The Warden, if the switch has been flicked up 43 times, then the leader can announce it without needing to flip the switch down the 44th time.
The question about the girl being kept in the cellar seems inappropriate given recent events.
The questions are not explained properly and the answers are not logical. There is a lot more possible answers e.g. why does the dwarf not take the umbrella everyday when he takes the lift if he knows that's the only way he can reach the buttons and he doesn't like asking others for a favour?
I don't agree with the explanation of the first problem. Even if one of the princes was wearing a black hat while the other two wore white hats, it would still be as fair as the three of them wearing white hats. With the one black hat - two white hats situation, there is no way for any of the princes to deduce their own hat color.
The solution for finding the fake coin can be done in 5 weighings by dividing the coins into two groups and weighing them against each other.
An alternative solution to the puzzle could be that she's a nurse.
The trick question about the average number of fingers on a person's left hand is incorrect. You should add the total number of fingers and divide by the number of people, not multiply by 0.
I appreciate the challenging puzzles you provide. I would love to see more puzzles that stimulate critical thinking.
The three persons can actually be the same. Suppose, I had an old friend who once saved my life and we haven't met for years. In the meantime, I had married another girl. So, finally now: I am an old man and for me all three are the same person.
The Emperor puzzle's answer is that only 4 prisoners die. The method involves dividing the bottles into sets and testing them with prisoners.
The solution to the fuse puzzle requires an extra condition on the burn rate for it to work as a logic puzzle. If the burning rate is random and unknown, the conclusion that lighting both ends will take 30 minutes may not hold true.
This puzzle is too hard for children; they would need help frequently.
Please post more lateral thinking puzzles.
The APPLE box can either be ORANGE or BOTH. The ORANGE box can either be APPLE or BOTH. The BOTH box can either be APPLE or ORANGE. This leads to a logical deduction based on the fruit picked.
The solution should be presented in a simpler way, as the current logic is not clearly understood.
The explanation about the equation involving fingers is interesting, but it could be clearer. It seems to suggest that multiplying by zero is unnecessary, which might confuse some readers.
The judge did not specify which woman he was speaking about. He could not punish the woman before him because it was her sister who had clearly murdered the husband.
The solution to problem 3 is incorrect. The second in command pirate realizes he won't get a majority in the vote if he joins the mutiny, so he will support the captain. This means the captain only needs one extra vote, which can be easily bought.
The riddle should avoid involving babies in harmful scenarios. Consider changing the wording to something less sensitive.
Regarding the woman having two sons born at the same time, another option is that she gave birth to twins on a day of daylight savings. She gave birth to one son at 1:55am, and the time had reset back to 1:00 by the time the second son was born.
The solution to the riddle is misleading as it assumes every person has five fingers. The second part of the line should be removed for clarity.
An alternative solution is that the fireman would not be playing cards at home; they would be at the fire station.
Zero should be excluded from the instruction to multiply the number of fingers on every person's left hands.
#1 is problematic as it is based on the unconfirmed supposition that being scared cures hiccups. I doubt the validity of putting a controversial subject such as hiccups in the facts section.
The stream question is flawed because it states you can paddle 7mph in a placid lake, not a stream.
The last puzzle is silly. The character could have found something buoyant, held onto the boat, or floated on his back.
Another solution for the fake coin puzzle is to split the 12 coins in half and weigh the lighter half. Then split the lighter half again and weigh to deduce which is the fake coin.
Another solution for the fake coin: we have 12 coins, split them in half and weigh the lighter of the 6 to find the fake coin. Split the lighter half and weigh again (3 vs 3). Take any 2 from the pile of 3 that weighs lighter to deduce which is the fake coin.
The solution involves using the numbers 5 and 9 only once, with 5 being greater and 9 being smaller.
The question in 'The Most Intelligent Prince' puzzle is not properly described, leading to confusion about the answer.
La respuesta del ejercicio de los trillizos está mal porque dos bebés no pueden nacer exactamente a la misma hora, siempre hay diferencia de hora aunque sea en segundos.
I suggest a different approach for the fake coin puzzle. Divide the coins into 6 pairs and weigh three pairs against three pairs. If they balance, discard those pairs and weigh the remaining pairs until you find the fake coin.
I believe I found a second solution to the fake coin puzzle.
The riddle about the man in the bar is intriguing. The solution involves understanding the context of the situation.
La respuesta la da el nombre.
The correct past tense of 'callar' in the puzzle is 'cayó', not 'calló'.
The question asks which one to choose for a ride, emphasizing the need to follow the rules of having only one passenger. The best answer should consider the context of the situation.
The seller question could be answered by anything possible. It seems like a poorly constructed question.
The question regarding the number of coins being the cube root of 3, such as 243, could be improved.
The third question of very difficult can have an easier solution: Put coins 1, 2, 3, 4, 5, 6 on one side and the rest on the other. Get the heavier side and repeat the step with 3 on each side. Now you have 3 coins; compare any two of them. If both weigh equal, then the third one is fake; otherwise, the chosen one will be heavier.
There is a flaw in the logic of puzzle #32 regarding the distribution of coins. The assumption that the coins are distributed in an unknown manner is incorrect.
The dice puzzle seems to lack a clear solution. Can you provide more details or clarify the problem?
The solution could also be 13/39 or 1/3, considering the bottom left quadrant.
The analysis must double count the square for BB born on a Tuesday because there are two scenarios within that square. The real answer is indeed 1/2.
This question has multiple answers. For example, if I ask one robot if he is a robot, then I know if he is a liar.
I have an alternate solution to the fake coin puzzle. Divide the 12 coins into four groups of 3 coins each and weigh them against each other to find the fake coin.
I think the fake coin puzzle can be solved in 5 weighings. Divide the coins into two parts and weigh them against each other to find the faulty coin.
In the first question, why not just give your keys to your friend and be with your perfect companion? You can always get your car later.
The mother was a psychopath, not an odd woman. I had to think about that for ages, thanks to that hint.
The Citrog solution does not seem valid. The battery is 10 km away with only one wire connected to it, making it impossible to light the bulb.
I enjoy the puzzles on your site. Please create more like the existing ones.
I think that the answer is '2'... as the number on dice is immaterial.
The probability calculation is incorrect; you need to count the intersection twice, making the probability 1/2.
This puzzle about the surgeon who cannot operate on the boy because he is his son is a classic riddle that highlights gender biases in assumptions.
This puzzle has too many assumptions and false statements, such as gender assumptions. It suggests that the dwarf 'has to walk'; why can't he carry the umbrella or another device when it's not raining?
For the fake coin puzzle, there is an alternative solution assuming coins 1, 2, 3, 4 and 5, 6, 7, 8 are not equal. You can take coins 1, 2, and 5, and balance them against coins 3, 4, and 6.
About the fake coin puzzle, I would divide the coins into 6 pairs and weigh them against each other. If the two pairs weigh the same, I would discard them and continue with the next pairs.
I believe I have a simpler solution for the fake coin puzzle. Weigh 6 coins against 6 coins. Set aside the heavier set and split the lighter set into 3 and 3. Weigh any 2 of the remaining 3 coins.
I believe I found a second solution to the fake coin puzzle. Can I share it with you?
The explanation of the abrupt stop to laughter in the puzzle makes no sense, and there is no mention of a present being dropped on any of the philosophers' heads.
I would like to offer an easier way to solve the fake coin puzzle. We have 12 coins, one of which is imperceptibly different. Split all coins into 6 pairs and weigh them against each other.
The puzzles are silly and the answers make no sense.
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