You Light Up for Life

A warden meets with 23 prisoners. He tells them the following:

  • Each prisoner will be placed into a room numbered 1-23. Each will be alone in the room, which will be soundproof, lightproof, etc. In other words, they will NOT be able to communicate with each other.
  • They will be allowed one planning session before they are taken to their rooms.
  • There is a special room, room 0. In this room are 2 switches, which can each be either UP or DOWN. They cannot be left in between, they are not linked in any way (so there are 4 possible states), and they are numbered 1 and 2. Their current positions are unknown.
  • One at a time, a prisoner will be brought into room 0. The prisoner MUST change one and only one switch. The prisoner is then returned to his cell.
  • At any time t, given some N>0, there exists a finite t_0 by which time every prisoner will have visited room 0 at least N times. (In other words, there is no fixed pattern to the order or frequency with which
    prisoners visit room 0, but at any given time, every prisoner is guaranteed to visit room 0 again. If you’re still confused by this statement, ignore it, and you should be ok).
  • At any time, any prisoner may declare that all 23 of them have been in room 0. If right, the prisoners go free. If wrong, they are all executed.

What initial strategy is 100% guaranteed to let all go free?

Communicated by Steven Johnson.

WARNING: of all the riddles, this one gave me the hardest time!

156 Comments

  1. Steven Miller on March 26, 2025 at 10:54 pm

    Nope. Email me at [email protected] for a hint



  2. Anonymous on March 26, 2025 at 10:53 pm

    109890? I’m a kid



  3. Steven Miller on March 24, 2025 at 5:30 am

    You can do it without assuming something like Chuck, they can just move the switches



  4. Steven Miller on January 24, 2024 at 9:33 pm

    That is included in the etc. you can solve without anything like that



  5. Penguin on January 24, 2024 at 9:17 pm

    It doesn’t say that there is no chalk in room 0. Or each prisoner originally has a tiny wedge of chalk with them. They mark the wall with a dot or dash, and when the prisoner that sees 23 markings can declare that they’ve all been in room 0.



  6. Steven Miller on March 14, 2019 at 5:16 am

    Please feel free to email me at [email protected] (the email address you used didn’t work). I think this is the hardest on the site



  7. cool kid 9000 on March 14, 2019 at 3:44 am

    this was very hard but i got the answer



  8. Steven Miller on August 10, 2017 at 11:41 pm

    email me: [email protected]



  9. bob on August 10, 2017 at 7:44 pm

    .



  10. Steven Miller on April 4, 2016 at 6:01 pm

    could visit a billion times!



  11. Cary on April 4, 2016 at 5:44 pm

    So, prisoner #1 could visit room 0 5 times before prisoner #5 goes in at all?



  12. Steven Miller on February 15, 2016 at 5:59 pm

    Nope. ..s



  13. Todd on February 15, 2016 at 5:53 pm

    does the first prisoner know he is the first prisoner to enter?



  14. Steven Miller on February 15, 2016 at 5:34 pm

    Nope. ..s



  15. Todd on February 15, 2016 at 5:31 pm

    can a prisoner flip a switch twice before leaving, IE flip the right switch one direction then the same switch back to its original position in one visit?



  16. Steven Miller on August 16, 2015 at 4:57 pm

    Yep. Or a million…..



  17. Todd on August 16, 2015 at 4:54 pm

    So a prisoner in say cell one, could be put in cell 0 10 times before the prisoner from cell 2 is put in cell 0 once?



  18. Steven Miller on May 4, 2015 at 8:47 am

    email to you bounces — please email me at sjm1 AT williams.edu



  19. Steven Miller on January 3, 2014 at 7:14 am

    I tried emailing you but it bounced. I worked with some students and we used WordPress.



  20. Agen Sbobet Online Terpercaya on January 3, 2014 at 1:02 am

    Cool blog! Is your theme custom made or did you download it
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  21. Steven Miller on October 7, 2013 at 1:30 pm

    your email address didn’t work. email me (sjm1 AT williams.edu) for a hint



  22. leslie on October 7, 2013 at 1:12 pm

    how a hard question sir . may i know the correct answer so that i’am going to present that as my project in mathematics ? please ?



  23. Steven Miller on September 24, 2013 at 1:06 am

    Nope, no role.



  24. nikos on September 23, 2013 at 9:19 pm

    does the fact that 23 is a prime number play any role for the solution?



  25. http://www.thethrifters.net/ on July 15, 2013 at 3:37 am

    Hello it’s me, I am also visiting this web page regularly, this web page is really nice and the users are really sharing nice thoughts.



  26. Steven Miller on March 28, 2013 at 8:05 pm

    email to you bounced — please email me at sjm1 AT williams.edu



  27. kelly on March 28, 2013 at 7:28 pm

    whats the answer?



  28. Steven Miller on March 18, 2013 at 7:57 am

    tried emailing you but emails bounced — email me at sjm1 AT williams.edu



  29. KuroganX on March 18, 2013 at 3:58 am

    Hmm. I am really quite confused. I just found this website and have amused me with interesting questions. may I please know the solution and explain how to get the answer. Thx.



  30. Steven Miller on December 25, 2012 at 1:18 pm

    email@email: essentially correct, email me at sjm1 AT williams.edu



  31. Steven Miller on December 24, 2012 at 1:38 pm

    [email protected]: you’re somewhat on the right track, but this email is not a valid account — email me at sjm1 AT williams.edu to chat



  32. Steven Miller on December 6, 2012 at 7:57 pm

    Cocobread: almost – email me at sjm1 AT williams.edu to discuss



  33. Steven Miller on October 23, 2012 at 1:18 am

    email me — sjm1 AT williams.edu



  34. john on October 22, 2012 at 7:35 pm

    how do you account for the unknown initial condition? (it could start up and you will mistakingly think 23 have been in when it was 22)



  35. Steven Miller on October 19, 2012 at 1:46 am

    Anonymous: well done — email sjm1 AT williams.edu



  36. Anonymous on October 18, 2012 at 3:45 pm

    I aced this one. There is a dummy switch. Im the only one allowed to turn the real switch down. Everyone else can turn the real switch only up, and only once. And i count.



  37. Steven Miller on September 18, 2012 at 8:56 pm

    sending hint



  38. john on September 18, 2012 at 10:53 am

    what is the answer my mind is pretty beaten up……



  39. Steven Miller on September 14, 2012 at 3:04 am

    sent — one of my favorites, and in my mind the hardest on the site



  40. robert on September 14, 2012 at 1:11 am

    Can I have the solution? This is an awesome riddle



  41. Steven Miller on July 31, 2012 at 5:56 am

    sure //s (sjm1 AT williams.edu)



  42. Catherine Bradley on July 31, 2012 at 1:41 am

    May I please have the solution? Thank you.



  43. Steven Miller on July 26, 2012 at 6:12 am

    Derek: you missed something — email me at sjm1 AT williams.edu to chat



  44. mark on July 20, 2012 at 4:51 am

    wow, by far your hardest riddle. spent a 3.5 hour car drive thinking about it and still nothing. any chance i could get the solution? i am truly stumped



  45. Steven Miller on July 17, 2012 at 3:45 am

    ok == sent some info on my soln .s



  46. Dale on July 17, 2012 at 1:34 am

    I am intrigued by your “You Light Up for Life” riddle. I think I may have an answer, but I was wanting to know if you would send me the answer to the riddle.



  47. Steven Miller on June 24, 2012 at 3:15 am

    ok: hint/soln sent



  48. Niherdh on June 23, 2012 at 4:34 pm

    Pls send me the answer. I couldn’t get it ;-(



  49. Steven Miller on June 22, 2012 at 2:44 am

    Daniel: close. email me at sjm1 AT williams.edu



  50. Steven Miller on June 22, 2012 at 2:44 am

    hint/soln sent



  51. Grinberg on June 22, 2012 at 2:10 am

    i spent 2 four hour shifts thinking and i got ….NADA…:)



  52. Steven Miller on June 22, 2012 at 2:00 am

    sure



  53. Grinberg on June 22, 2012 at 1:46 am

    the more i think about it the harder it gets…..can you please send me the solution?



  54. Steven Miller on June 21, 2012 at 4:05 am

    these are immortal people — it could be a billion years before one of the people visits!



  55. Viet on June 21, 2012 at 3:57 am

    How many times a day do these prisoners visit the room? Why not just wait 3 years? Their lives are on the line! oR wait 10 years. how old are these prisoners? If they are all almost dying from age might as well wait 1 year then anybody can declare. cos they are going to die in prison if they dont get executed anyway!



  56. Steven Miller on June 6, 2012 at 3:30 pm

    Alex Irby: Correct. Please include an email so I can comment on your solns, or email me at sjm1 AT williams.edu //s



  57. Hildy on May 10, 2012 at 3:54 pm

    Any chance you’ll give us the answer? 🙂



  58. Steven Miller on May 9, 2012 at 5:47 am

    sure, coming over. I think this is the hardest one on the site. /s



  59. Dylan on May 8, 2012 at 9:20 pm

    While a prisoner can only change one switch is it possible for the prisoner to change that one switch more than once? Say change it then change it back?
    Also could you please send me the answer for this riddle? Hard to concentrate on studying when you are stuck in riddles!



  60. Steven Miller on April 25, 2012 at 5:20 pm

    you can visit a room many times in any order. that said, no guarantee that every prisoner will visit once prior to someone visiting twice.



  61. Phil on April 25, 2012 at 1:59 pm

    Question: Can a particular prisoner visit room zero more than once prior to any other prisoner not entering the room? Or more specifically, is there any guarentee that every prisoner will visit room zero prior to any other prisoner visiting room zero twice?



  62. Steven Miller on April 17, 2012 at 2:11 pm

    I still believe this is the hardest on the site. I’ll let the comment go through as you’re the proposer. //s



  63. Steven Miller on April 17, 2012 at 2:09 pm

    ok, hint/soln sent



  64. ncn on April 17, 2012 at 12:06 pm

    i am clueless…



  65. Steve Johnson on April 17, 2012 at 5:29 am

    Wow. Glad to see my problem is still generating a buzz!

    Here are hints I typically give if people are trying to solve it but don’t just want the answer. You can either let this comment through or reject it and add hints in the usual sense in case people want them separately.

    Hint 1: Start out by removing some assumptions, then seeing if you can work them back in.

    Hint 2: Having two switches and being forced to flip one is not very different from having one switch and the option whether or not to flip it.

    Hint 3: (More explicit stating of hint 1): Start with the case of 2 prisoners, 1 switch, optional whether or not to flip it, known to start down. This should be fairly simple. Then, attack each assumption one at a time. What modifications need to be made to generalize the number of prisoners? What modifications need to be made to account for having 2 switches and being forced to flip one? What modifications need to be made to get around the initial configuration being unknown?

    Good luck!



  66. Steven Miller on April 2, 2012 at 12:46 am

    correct //s



  67. Viktor on April 1, 2012 at 7:40 pm

    So, the prisoners who doesnt get chosen to go to room 0, cant see when someone is picked, or see how long the one who does, stay there?



  68. Steven Miller on March 27, 2012 at 6:31 pm

    If memory serves, the problem with your soln was you don’t know the initial configuration of the lightswitches, or who starts….



  69. sauravshakya on March 27, 2012 at 11:39 am

    wasnt my solution correct



  70. Steven Miller on March 15, 2012 at 2:34 am

    To anonymouns (is it that every person will be told to ….): correct, well done! not posting as its’ the soln



  71. Steven Miller on March 15, 2012 at 2:32 am

    I’m not sure. I struggled for awhile on this one, finally solved it by taking a 50 minute walk



  72. nikki on March 15, 2012 at 2:03 am

    what is the fastest time a person finished this riddle?



  73. Albert on March 14, 2012 at 6:36 am

    Can you please email me the solution:D



  74. Steven Miller on March 13, 2012 at 4:07 am

    sadly, yes — one of the many things that makes this problem so hard! email me at sjm1 AT williams.edu if you want more of a hint / the soln



  75. Mark on March 12, 2012 at 2:12 pm

    Can I please get some clarification on this problem?
    Is it possible for the same prisoner to be brought to room 0 twice in a row, without any other prisoners visiting the room? If so I can’t see a solution that guarantees success.



  76. Steven Miller on March 11, 2012 at 1:44 am

    almost — you’d be correct if you knew the initial configuration of switches.



  77. Steven Miller on March 6, 2012 at 2:08 am

    nope — this is the hardest one by far on the page! :]



  78. confused on March 6, 2012 at 1:20 am

    before i continue trying to crack my brain on this one…is there some sort of non mathematical “trick” involved?



  79. Steven Miller on February 26, 2012 at 6:01 am

    Stan: you’re close; email me ([email protected]) if you want a hint / my soln. //s



  80. Stan on February 26, 2012 at 4:42 am

    Well I guess that everyone would flip the first switch but then the 22nd guy would flip the 2nd switch so the next guy would know that he was 23rd, i just cant figure out how he would figure out he was the 22nd guy to go in so he could flip the 2nd switch



  81. Steven Miller on February 22, 2012 at 12:45 pm

    same here — a fun problem



  82. Anonymous on February 22, 2012 at 2:16 am

    Greetings From Canada I agree with you 97% I have some reservations, but on the whole as far as I am concerned your views are good enough



  83. Steven Miller on February 21, 2012 at 2:00 pm

    Well done! ..s



  84. sam on February 21, 2012 at 9:19 am

    Sent you the solution



  85. Steven Miller on February 18, 2012 at 9:54 pm

    will do — I think this is the hardest. //s



  86. Kristina on February 18, 2012 at 3:36 am

    I was thinking about this really hard but… why not just ask the Warden?
    But that would defeat the purpose of the riddle… Hm. I’d say what I think, but I’m sure I’m wrong. Does my email show up for you? If so, please send me the answer. :/ I agree, this is probably one of the hardest, if not the hardest, one on the sight.



  87. Steven Miller on February 17, 2012 at 4:44 pm

    Glad you’re enjoying the page. It’s interesting seeing how much the traffic is increasing — at this rate in a year or two I’ll need to have someone helping!



  88. proxy for senuke on February 17, 2012 at 8:07 am

    Youre so cool! I dont suppose Ive read anything like this prior to. So nice to obtain somebody with some original thoughts on this topic. realy thank you for starting this up. this web site is some thing that is needed on the web, someone using a small originality. beneficial job for bringing some thing new to the web!



  89. Steven Miller on February 7, 2012 at 5:45 pm

    I believe it’s the hardest one on the site — email me at [email protected] if you want a hint. //s



  90. Anonymous on February 7, 2012 at 2:40 pm

    this is hard



  91. Steven Miller on February 2, 2012 at 2:22 am

    sure //s



  92. kevin k on February 2, 2012 at 2:20 am

    will you pls. e-mail me the sol.?



  93. Tyson on February 2, 2012 at 12:35 am

    can you email me the answer



  94. Steven Miller on February 1, 2012 at 9:11 pm

    Hmm — will you be honest and admit to them that you had to look it up? If you’re willing to do that I’ll send the answer to “you light up for life”….



  95. kenlei on February 1, 2012 at 4:51 pm

    please send me the answer, I am trying to stump some coworkers and the rules of the office is that you must know the answer first:(



  96. Steven Miller on January 31, 2012 at 2:31 pm

    Sure — this is the hardest one on the site (in my opinion). //s



  97. Khairul on January 31, 2012 at 9:10 am

    Can you please email me the solution to this email?
    I am very interested about it!
    Thank you in advance.



  98. Steven Miller on January 31, 2012 at 4:11 am

    Sure — this is I believe the hardest one on the site.



  99. Darel Terral on January 30, 2012 at 6:01 pm

    Can you email the answer to this email. Students checking my answer.
    Thanks



  100. Steven Miller on January 27, 2012 at 2:44 am

    this is the hardest (in my mind) on the site — will email.



  101. John Franks on January 26, 2012 at 9:56 pm

    Hey Steve,

    My Friend and me have been trying to figure this problem out for a couple weeks now and I was hoping that you could give us the answer and the math on how to get the answer.

    Thanks



  102. Steven Miller on January 18, 2012 at 7:46 pm

    Typically the numbers matter little, if at all. You could do 2012 prisoners; people often use the year to highlight the fact that the number doesn’t matter. Glad you’re enjoying the site. //s



  103. Nayra on January 18, 2012 at 4:26 pm

    Gosh! It gets more complicated the more you think about it.
    I need a small hint… Does the riddle only work with 23 prisoners? If they can come in and out as many times as you want, the number is irrelevant, is it? Could it be done with, say, 15 prisoners? Could it be done with 46?
    By the way, great website! I can’t believe I hadn’t found it before!



  104. Steven Miller on January 13, 2012 at 4:23 pm

    Sure. Let me know if you didn’t get it (email me at [email protected])



  105. Tony on January 13, 2012 at 4:19 pm

    can you email me the anwser



  106. Steven Miller on January 6, 2012 at 5:19 am

    not sure — what does it mean to double?



  107. Bren on January 5, 2012 at 10:13 pm

    can i double my chanses?



  108. Steven Miller on December 17, 2011 at 2:24 pm

    email to y ou isn’t working — email me at [email protected]



  109. subhas on December 17, 2011 at 10:21 am

    can you send me the solution/



  110. Steven Miller on December 12, 2011 at 2:33 am

    Andrew: essentially correct (which is why I’m not posting), well done!



  111. Adam on December 7, 2011 at 8:08 pm

    Yay for solving 🙂



  112. Steven Miller on December 7, 2011 at 2:11 pm

    Just send me an email ([email protected])



  113. Adam on December 7, 2011 at 3:26 am

    I believe I know the answer. How do I find out if I am correct?



  114. Steven Miller on December 2, 2011 at 3:14 pm

    yes. //s



  115. Ben on December 2, 2011 at 2:31 pm

    Let me get this straight: we have 22 people with one leader, and they move one switch unless there is some special circumstance, in which case they move the other. Am I right?



  116. Steven Miller on December 1, 2011 at 6:40 pm

    locked in their rooms, and *I* get to decide the order. I can be as mean as I want. I can listen to their plotting. If I want, I can have prisoners 2 and 4 alternate one gazillion times in a row before someone else enters!



  117. kareem on December 1, 2011 at 5:39 pm

    One more question, are the prisoners locked in each of their rooms? who decides the order in which they get to leave the room to go to room 0? is this something they could decide upon during the initial meeting?



  118. Steven Miller on December 1, 2011 at 2:03 am

    sadly, nope. there is a solution without having a starting state that is known to all



  119. kareem on November 30, 2011 at 10:35 pm

    “Their current positions are unknown” regarding the switches, thats what’s written. If we say that they had their initial planning meeting in room 0, could we say that they arranged the switches in a certain way? both up, or both down, for example?



  120. Steven Miller on November 26, 2011 at 1:34 am

    they were gunned down — warden is no fool and had the room bugged!



  121. Some guy on November 24, 2011 at 10:16 pm

    Durring the planning session, they all team up to kill the warden and escape the prison. 😛 (I’m just joking)



  122. Steven Miller on November 24, 2011 at 1:56 pm

    This is the hardest by far on the site!



  123. Some guy on November 24, 2011 at 6:34 am

    Wow. This riddle is extremely difficult! I don’t understand how someone with an IQ any lower than 500 can figure out the answer to this! It’s going to take me a while to figure out the answer to this one. (If I ever do figure it out)



  124. Josh H on November 23, 2011 at 3:25 am

    One more question, can they visit Room 0 before the start of the game, like for their meeting session



  125. Steven Miller on October 31, 2011 at 12:21 pm

    That’s a good hint. So you can say there is a special person, and all else are equal, and there is a special switch that is used as a special switch if possible. This is a really hard riddle (I think the hardest on the site).



  126. Mike on October 31, 2011 at 11:40 am

    Second innocent hint??!
    What if the two switches are assigned separate functions???

    [I should say, I have a solution, and being a complete riddle fanatic myself, I am trying to deliver some useful hints to any other fanatic, that won’t be utter spoilers].



  127. Steven Miller on October 30, 2011 at 10:17 pm

    that’s the right way to start — one prisoner is special.



  128. Mike on October 30, 2011 at 1:10 pm

    Innocent hint?!!

    What if the prisoners, nominate a leader between them?



  129. Steven Miller on October 30, 2011 at 1:14 am

    this is the hardest one on the site — if you want a hint/soln, email me at [email protected]



  130. mukuro on October 29, 2011 at 4:30 am

    SO HARD
    SO COMPLICATED!
    AHAHAHAHAHA
    i can’t answer it, ahahahaha



  131. Steven Miller on October 28, 2011 at 7:33 pm

    ok, but they still need to agree on a strategy!



  132. Jon on October 28, 2011 at 3:29 pm

    They hold their initial meeting/planning session in room 0.



  133. Steven Miller on October 9, 2011 at 1:53 am

    To: Anonymous: Submitted on 2011/10/08 at 7:22 pm: First person can only flip one switch, people can only flip switches cannot use anything to write their names.



  134. Steven Miller on October 5, 2011 at 2:09 am

    this is almost surely the hardest problem on the site

    if your teacher gets an answer, I’m happy to check.



  135. Olivia on October 5, 2011 at 1:46 am

    I gave this problem to my math teacher, could i have the solution o i can check what she gets? [email protected]



  136. Steven Miller on September 26, 2011 at 2:41 am

    Yes, can keep coming and coming and coming…..



  137. quess on September 25, 2011 at 11:41 am

    I need to know one more thing from this:

    If no-one declares to be 23th person. Does prisoners continue visiting in roomto end of a world?

    So is there some timelimit or other limit?



  138. Steven Miller on September 10, 2011 at 3:42 am

    Yep — it’s these issues that make it so hard!



  139. chris mercer on September 9, 2011 at 6:48 am

    Dear Steve,
    this is a tough one.
    do i understand the rules correctly, that 1) the first prisoner to get to room 0 does not know he is the first? and 2) theoretically it could happen that let us say prisoner 7 might only get to room 0 for the first time after one thousand other room-0-visits by the other prisoners?
    regards
    chris



  140. Steven Miller on September 4, 2011 at 2:56 pm

    Thanks for your kind words. It’s a lot of fun chatting with people about math. I’m working on a teacher’s corner with solutions and putting the riddles in context. If you want to be added to the list (or want to help), let me know. //s



  141. Trinity Simkin on September 4, 2011 at 9:19 am

    I would like to get across my admiration for your kindness supporting folks that ought to have guidance on in this subject matter. Your individual dedication to obtaining the solution throughout had become particularly advantageous and has encouraged regular individuals just like me to obtain their pursuits. Your insightful data indicates a good deal to me and substantially more to my fellow workers. Thanks a whole lot; from all of us.



  142. Steven Miller on August 23, 2011 at 6:05 pm

    Right, but you need something that will work in general….



  143. Joona on August 22, 2011 at 8:35 am

    I have kind of an ansrew. It is not full-proof but i will work with some ways.

    When starting postions are followings: 1,2 1,1 2,2 2,1 (1 = up 2 = down)
    Every time when some one moves one of those, there will be added +1 or -1 to value ( so i mean if position is 1,2 next value is 1+1,2 or 1, 2-1. ) If u count it longer u notice that in 23 adds ending values always uneven son 1,3,5,7 and so on. U can transfer that to switch postions so uneven value is always 1,1 or 2,2. So if swich are in postion 1,2 or 2,1 it cant be ended yet. I cant say more yet but i will correct this ansrew later. There is now half of the values deleted. But it will not solve this problem yet.



  144. Steven Miller on July 28, 2011 at 5:10 pm

    no



  145. Shane on July 28, 2011 at 4:25 pm

    Is it required that each prisoner must visit at least once before any prisoner visits a 2nd time?



  146. Steven Miller on July 6, 2011 at 3:13 am

    Really hard to give a hint for this one; try making one special.



  147. brenda on July 6, 2011 at 3:00 am

    can you give a hint? so confusing!



  148. Soundproofing on May 24, 2011 at 5:29 am

    Hello Steven. Nice. I will try it.Thanks.



  149. Steven Miller on May 3, 2011 at 1:35 am

    Without a doubt, this was the hardest for me to solve. It took me a 50 minute walk away from everything to see how to tackle it. If you want a hint, email me at sjm1 AT williams.edu.



  150. Anonymous on May 2, 2011 at 7:59 pm

    My head hurt solving this



  151. Steven Miller on April 7, 2011 at 1:47 pm

    Sorry to hear that; however, if you have some feline blood you might still have 6 lives in reserve! For me, riddles like this are great at getting me to think in radically new ways.



  152. jfjksdfkl on April 6, 2011 at 4:40 pm

    This game has no purpose. I would have been executed 3 times!!



  153. Steven Miller on March 23, 2011 at 5:33 pm

    I don’t see how to do it with one if each person must flip a switch every time they enter. Could you email me (sjm1 AT williams.edu) your solution?



  154. uri on March 23, 2011 at 5:22 pm

    one switch is enough to solve this one.



  155. Steven Miller on March 23, 2011 at 2:00 pm

    In my mind, this is the hardest problem on the site. It took me a long time to see the right way to look at it (I think it was in a 50 minute walk from my wife’s office back to our house).



  156. Jaap on March 23, 2011 at 1:39 pm

    A very ingenious problem indeed. It took my colleagues and me the best part of a lunch break to solve, after two of us had spent some thought on it overnight. Once you know the solution, all parts of the problem make perfect sense.



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