IQPOTD - IQ Problem Of The Day
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Suppose you have twelve balls, they all look the same. One of them has a different weight while the rest are identical. You are given a scale that you can place two subgroups of the balls on it to see which subgroup is heavier. You can decide how many balls goes into each subgroup. You don't know the exact weight of each subgroup, but you are allowed to remember which balls belong to which subgroup (the heavier one or the lighter one). The problem is that you have to find a way to single out the ball that has a different weight by using the scale only three times. No formal education or special knowledge is required, but it does take some creative thinking. There are multiple answers, by the way. A. If you can do this within one hour, you are extremely smart, maybe one in a million. B. If you can do this within one day, you are still pretty smart. C. If you did it in less than 10 minutes, then you are just lucky or you have done a similar problem before. ;)[
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Is this supposed to be a hard problem? :~ It's easily solved with a classical divide and conquer algorithm. Either I'm extremely smart, or I've studied basic computer science and/or discrete mathematics. :rolleyes: -- Ich bin Joachim von Hassel, und ich bin Pilot der Bundeswehr. Welle: Erdball - F104-G Starfighter
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Is this supposed to be a hard problem? :~ It's easily solved with a classical divide and conquer algorithm. Either I'm extremely smart, or I've studied basic computer science and/or discrete mathematics. :rolleyes: -- Ich bin Joachim von Hassel, und ich bin Pilot der Bundeswehr. Welle: Erdball - F104-G Starfighter
The answer given by Denervers --- and probably the answer everyone is thinking of --- assumes that you know if the odd ball is lighter or heavier than the others. If you only know that it has a different weight, but don't know if it is heavier or lighter, then the problem is a lot harder. John Carson "I believe in an America where the separation of church and state is absolute--where no Catholic prelate would tell the President (should he be Catholic) how to act, and no Protestant minister would tell his parishoners for whom to vote ... and where no man is denied public office merely because his religion differs from the President who might appoint him or the people who might elect him. - John F. Kennedy
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The answer given by Denervers --- and probably the answer everyone is thinking of --- assumes that you know if the odd ball is lighter or heavier than the others. If you only know that it has a different weight, but don't know if it is heavier or lighter, then the problem is a lot harder. John Carson "I believe in an America where the separation of church and state is absolute--where no Catholic prelate would tell the President (should he be Catholic) how to act, and no Protestant minister would tell his parishoners for whom to vote ... and where no man is denied public office merely because his religion differs from the President who might appoint him or the people who might elect him. - John F. Kennedy
John Carson wrote: If you only know that it has a different weight, but don't know if it is heavier or lighter, then the problem is a lot harder. Then it's probably impossible to tell. I'm not sure, but I have a pretty good hunch that it is. :) -- Ich bin Joachim von Hassel, und ich bin Pilot der Bundeswehr. Welle: Erdball - F104-G Starfighter
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1 - take the 12 balls, split them in 6 balls groups on each side of the scale, pick the 6 balls that are heaviest. 2- split in 3 balls group, repeat the same experiment, pick the heiviest group. 3- you now have 3 balls, pick any 2 to weight them on each side and pick the heaviest, if they are the same weight, the one you have in your hands in the right one. took me 5 minutes (honest) oh.. I get it, it's not a real IQ test. dang !
Denevers wrote: oh.. I get it, it's not a real IQ test. dang ! Did I mention you need to find the correct answer? P.S. In order to find the correct answer, you need to have a correct understanding of the problem. ;)[
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John Carson wrote: If you only know that it has a different weight, but don't know if it is heavier or lighter, then the problem is a lot harder. Then it's probably impossible to tell. I'm not sure, but I have a pretty good hunch that it is. :) -- Ich bin Joachim von Hassel, und ich bin Pilot der Bundeswehr. Welle: Erdball - F104-G Starfighter
Jörgen Sigvardsson wrote: Then it's probably impossible to tell. I don't blame you. I actually wasted time trying to prove that it is impossible. :-D[
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I don't believe it. For once, someone posted a puzzle, and I figured it out right away.
Software Zen:
delete this;Gary R. Wheeler wrote: I don't believe it. For once, someone posted a puzzle, and I figured it out right away. I tend to believe you are one of the smartest people, because nobody can be that lucky. :-D By the way, you need to hand in your homework to get credit for it. ;)[
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Hey, if I had 12 balls I wouldn't need to be smart!:laugh:
The only way of discovering the limits of the possible is to venture a little past them into the impossible.--Arthur C. Clark
Toasty0 wrote: Hey, if I had 12 balls I wouldn't need to be smart! I agree. But it is still a hard problem to figure out which one of your balls has a different weight, isn't it? :-D Sorry, I can't resist it.[
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For those who like an even bigger challenge, try the same problem with 120 balls and using the scale only 5 times... Or with 64570080 balls and using the scale 17 times... (You'd need a pretty big scale, I admit, but it can be done!) Jeffrey
Everything should be as simple as possible, but not simpler.
-- Albert Einstein http://www.extremeoptimization.com/Jeffrey Sax wrote: For those who like an even bigger challenge I don't know about that, unless there is a different kind of trick involved. Walking 50 miles is harder than walking 5 mile, but it is hardly any more creative.[
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Suppose you have twelve balls, they all look the same. One of them has a different weight while the rest are identical. You are given a scale that you can place two subgroups of the balls on it to see which subgroup is heavier. You can decide how many balls goes into each subgroup. You don't know the exact weight of each subgroup, but you are allowed to remember which balls belong to which subgroup (the heavier one or the lighter one). The problem is that you have to find a way to single out the ball that has a different weight by using the scale only three times. No formal education or special knowledge is required, but it does take some creative thinking. There are multiple answers, by the way. A. If you can do this within one hour, you are extremely smart, maybe one in a million. B. If you can do this within one day, you are still pretty smart. C. If you did it in less than 10 minutes, then you are just lucky or you have done a similar problem before. ;)[
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Count me extremely lucky. I did it in less than ten seconds, no joke.
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The answer given by Denervers --- and probably the answer everyone is thinking of --- assumes that you know if the odd ball is lighter or heavier than the others. If you only know that it has a different weight, but don't know if it is heavier or lighter, then the problem is a lot harder. John Carson "I believe in an America where the separation of church and state is absolute--where no Catholic prelate would tell the President (should he be Catholic) how to act, and no Protestant minister would tell his parishoners for whom to vote ... and where no man is denied public office merely because his religion differs from the President who might appoint him or the people who might elect him. - John F. Kennedy
Right, I've a strong hunch that it would always take four tests then.
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Denevers wrote: oh.. I get it, it's not a real IQ test. dang ! Did I mention you need to find the correct answer? P.S. In order to find the correct answer, you need to have a correct understanding of the problem. ;)[
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Suppose you have twelve balls, they all look the same. One of them has a different weight while the rest are identical. You are given a scale that you can place two subgroups of the balls on it to see which subgroup is heavier. You can decide how many balls goes into each subgroup. You don't know the exact weight of each subgroup, but you are allowed to remember which balls belong to which subgroup (the heavier one or the lighter one). The problem is that you have to find a way to single out the ball that has a different weight by using the scale only three times. No formal education or special knowledge is required, but it does take some creative thinking. There are multiple answers, by the way. A. If you can do this within one hour, you are extremely smart, maybe one in a million. B. If you can do this within one day, you are still pretty smart. C. If you did it in less than 10 minutes, then you are just lucky or you have done a similar problem before. ;)[
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Solution 1: Sheer luck pick one at random. You don't know which one it is either right ? Solution 2: Psychic. hire a medium, use tea leaves, cards, pendulum. Solution 3: Sneaky you: oh, Xiangyang Liu, look behind you ! (replace the balls by your own 12 balls) Solution 4: Behavioral tell the balls to team up for some game.. of course, the one who is different will be picked up the last. Solution 5: Systematic Compare them all, using 12! (12 factorial -- I'm not surprised), change you name at every 3 trials. Solution 6: Bribery Mumble "I have a deal you can't refuse" with an faux italian accent. Everybody has its price. Solution 7: Google type "which ball has a different weight" in google and sort the 738000 answers. If it's not in Google, it does not exist. Solution 8: Practical Bring you own scale ! ;P
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Solution 1: Sheer luck pick one at random. You don't know which one it is either right ? Solution 2: Psychic. hire a medium, use tea leaves, cards, pendulum. Solution 3: Sneaky you: oh, Xiangyang Liu, look behind you ! (replace the balls by your own 12 balls) Solution 4: Behavioral tell the balls to team up for some game.. of course, the one who is different will be picked up the last. Solution 5: Systematic Compare them all, using 12! (12 factorial -- I'm not surprised), change you name at every 3 trials. Solution 6: Bribery Mumble "I have a deal you can't refuse" with an faux italian accent. Everybody has its price. Solution 7: Google type "which ball has a different weight" in google and sort the 738000 answers. If it's not in Google, it does not exist. Solution 8: Practical Bring you own scale ! ;P
Denevers wrote: Solution 7: Google type "which ball has a different weight" in google and sort the 738000 answers. If it's not in Google, it does not exist. Choose your keywords a bit more carefully and you should find a solution on the first page. John Carson "I believe in an America where the separation of church and state is absolute--where no Catholic prelate would tell the President (should he be Catholic) how to act, and no Protestant minister would tell his parishoners for whom to vote ... and where no man is denied public office merely because his religion differs from the President who might appoint him or the people who might elect him. - John F. Kennedy
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Suppose you have twelve balls, they all look the same. One of them has a different weight while the rest are identical. You are given a scale that you can place two subgroups of the balls on it to see which subgroup is heavier. You can decide how many balls goes into each subgroup. You don't know the exact weight of each subgroup, but you are allowed to remember which balls belong to which subgroup (the heavier one or the lighter one). The problem is that you have to find a way to single out the ball that has a different weight by using the scale only three times. No formal education or special knowledge is required, but it does take some creative thinking. There are multiple answers, by the way. A. If you can do this within one hour, you are extremely smart, maybe one in a million. B. If you can do this within one day, you are still pretty smart. C. If you did it in less than 10 minutes, then you are just lucky or you have done a similar problem before. ;)[
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I've never seen a problem like this before, and it took me just over 3 minutes to solve it. Does that mean I'm one in a thousand-kagillion? :P Was fun though!
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Suppose you have twelve balls, they all look the same. One of them has a different weight while the rest are identical. You are given a scale that you can place two subgroups of the balls on it to see which subgroup is heavier. You can decide how many balls goes into each subgroup. You don't know the exact weight of each subgroup, but you are allowed to remember which balls belong to which subgroup (the heavier one or the lighter one). The problem is that you have to find a way to single out the ball that has a different weight by using the scale only three times. No formal education or special knowledge is required, but it does take some creative thinking. There are multiple answers, by the way. A. If you can do this within one hour, you are extremely smart, maybe one in a million. B. If you can do this within one day, you are still pretty smart. C. If you did it in less than 10 minutes, then you are just lucky or you have done a similar problem before. ;)[
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I've never seen a problem like this before, and it took me just over 3 minutes to solve it. Does that mean I'm one in a thousand-kagillion? :P You can also do the same with 18 or 27 balls and 3 uses of the scales. or if you were allowed 4 uses of the scales, you could have 54 or 81 balls. Was fun though! thanks!
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I've never seen a problem like this before, and it took me just over 3 minutes to solve it. Does that mean I'm one in a thousand-kagillion? :P You can also do the same with 18 or 27 balls and 3 uses of the scales. or if you were allowed 4 uses of the scales, you could have 54 or 81 balls. Was fun though! thanks!
pmartin wrote: I've never seen a problem like this before, and it took me just over 3 minutes to solve it. Does that mean I'm one in a thousand-kagillion? The answer is available on the web, but if you did get the answer by youself, I strongly recommend that you post the answer in a new thread in the lounge, I am sure some people will be interested in this. I will nominate you as "The Problem Solver on CP", Chris will probably write a script to automatically find all homework problems in various discussion forums and forward them to your mailbox. :-D[
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pmartin wrote: I've never seen a problem like this before, and it took me just over 3 minutes to solve it. Does that mean I'm one in a thousand-kagillion? The answer is available on the web, but if you did get the answer by youself, I strongly recommend that you post the answer in a new thread in the lounge, I am sure some people will be interested in this. I will nominate you as "The Problem Solver on CP", Chris will probably write a script to automatically find all homework problems in various discussion forums and forward them to your mailbox. :-D[
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Xiangyang Liu wrote: I will nominate you as "The Problem Solver on CP", Chris will probably write a script to automatically find all homework problems in various discussion forums and forward them to your mailbox. Hehehe. Do I get a medal, or a framed letter? I hope so! I'll even fly down to sydney to pick it up Chris, if you are listening! :P But yeah, I hadn't seen it before, and no, I didn't look for it on the web. So hooray for me. All your homework are belong to us! - Phil
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Xiangyang Liu wrote: I will nominate you as "The Problem Solver on CP", Chris will probably write a script to automatically find all homework problems in various discussion forums and forward them to your mailbox. Hehehe. Do I get a medal, or a framed letter? I hope so! I'll even fly down to sydney to pick it up Chris, if you are listening! :P But yeah, I hadn't seen it before, and no, I didn't look for it on the web. So hooray for me. All your homework are belong to us! - Phil
pmartin wrote: But yeah, I hadn't seen it before, and no, I didn't look for it on the web. So hooray for me. BTW, I would like to see your answer. Not that I don't trust you, but there is a slight chance your solution is incorrect. I do it myself all the time, I mean providing wrong answers to other people. :-D[
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