Bridge Over Troubled Students

There are four men who want to cross a bridge. They all begin on the same side. You have 17 minutes to get all of them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either one or two people, must have the flashlight with them. The flashlight must be walked back and forth, it cannot be thrown, etc. Each man walks at a different speed. A pair must walk together at the rate of the slower man

Man 1: 1 minute to cross
Man 2: 2 minutes to cross
Man 3: 5 minutes to cross
Man 4: 10 minutes to cross

For example, if Man 1 and Man 4 walk across first, 10 minutes have elapsed when they get to the other side of the bridge. If Man 4 returns with the flashlight, a total of 20 minutes have passed, and you have failed the mission.

Communicated by Jeff Miller.
bridge

66 Comments

  1. Steven Miller on April 20, 2018 at 12:24 pm

    so glad to hear. once you send a correct soln I’m happy to send the password to the student/teacher corner, which has supplemental material.



  2. James Livingston on April 20, 2018 at 12:03 pm

    I love using these riddles with my students.



  3. Steven Miller on June 8, 2017 at 2:31 pm

    Yes. I keep it hidden as it’s valuable to work on the problems first and if there are posted solutions often the temptation is too great…. If you are a teacher, once you solve a problem I’m happy to give you access to the student/teacher corner, which has discussions on how to use these in classes.



  4. Clark on June 8, 2017 at 2:29 pm

    Is there an answer key? Thanks



  5. Steven Miller on March 13, 2017 at 7:27 am

    Once you solve a problem email me at [email protected] (or email me for a hint). I’ll then provide, if you wish, the password for the student/teacher corner, which has a lot of aids to use these in classes.



  6. Cher on March 3, 2017 at 7:15 pm

    Hi I’m a homeschool teacher. Amy I please have the solution to this problem?



  7. Steven Miller on February 7, 2017 at 5:56 pm

    email me at [email protected] for a hint



  8. tasiras on February 7, 2017 at 3:42 pm

    what is the answer



  9. Steven Miller on December 25, 2016 at 3:10 pm

    Please write in English



  10. Steven Miller on July 3, 2016 at 4:49 am

    Sure — drop me an email ([email protected]), but I think this comment was meant for a different riddle.



  11. Prakash Rajendran on July 1, 2016 at 9:03 pm

    Thank you for the problem and its solution which I noticed on the comment column. Could you please tell me how to connect to binomial expansion or any algebraic expression.
    Thanks



  12. Steven Miller on April 4, 2016 at 6:00 pm

    you eventually realize there are a few very popular problems with many
    variants!



  13. Jayashree on April 4, 2016 at 4:45 pm

    We have a classic problem for children similar to this, in India. It goes like this- a man has to ferry a tiger, a goat and a bundle of hay across a river. He can ferry only one at a time, or the boat would sink. He can’t leave the goat and the tiger together and take the hay, or the tiger would eat the goat. Also , the hay and the goat can’t be left in a bank alone. How would he ferry all three? Children love this one.



  14. Steven Miller on January 26, 2016 at 10:18 pm

    yes



  15. anonymous on January 26, 2016 at 9:58 pm

    do you also add the time it takes for man x to get back across w/the flashlight?



  16. Steven Miller on August 24, 2015 at 1:25 pm

    yes



  17. Dwayne on August 24, 2015 at 1:22 pm

    Do all four have to end on the other side?



  18. Steven Miller on August 5, 2014 at 1:54 pm

    [email protected]: your email doesn’t work, your soln is correct.



  19. Steven Miller on May 28, 2014 at 3:11 pm

    nope — must go all the way



  20. John Jackson on May 27, 2014 at 2:52 pm

    We are really enjoying these but we are stuck here, is there a way to go part of the way across without the flashlight?



  21. Steven Miller on December 1, 2012 at 4:20 pm

    email to you bounced – email me at sjm1 AT williams.edu



  22. brenda on December 1, 2012 at 9:53 am

    plz tell me the ans i can’t figue it out and i want to ask my friends. i can’t ask them without knowing the ans



  23. Steven Miller on October 31, 2012 at 1:37 am

    anonymous: correct (sjm1 at williams.edu)



  24. Steven Miller on June 13, 2012 at 1:16 am

    I’ll send a hint



  25. Katie on June 12, 2012 at 1:29 pm

    This is really confusing!!!



  26. Steven Miller on June 6, 2012 at 4:27 am

    the flashlight is very weak and only works if you’re right next to it. //s



  27. Chris on June 6, 2012 at 3:37 am

    The key to this riddle (to me) is the following statement: ‘Any party who crosses, either one or two people, must have the flashlight with them’.

    Answer:

    First, 4 and 3 cross the bridge. Then, 4 takes the flashlight to go get 1. However, while 4 is crossing to get 1, 1 is walking towards 4 (this is allowed because the stem states that the flashlight must be with one of them, not that they must be right next to each other) and then 1 walks 4 towards the other end (i.e., towards 3). Once, 1 walks 4 to the other side of the bridge, he (man 1) takes the flashlight and walks towards 2, while 2 walks towards 1. When they meet, 2 and 1 walk towards the end of the bridge. This process takes roughly 13 minutes.



  28. Steven Miller on June 1, 2012 at 7:09 pm

    Ritish: correct (and correct on other one); not posting as these are the solns



  29. Steven Miller on May 6, 2012 at 11:56 am

    let me send a hint first ..s



  30. Marc on May 6, 2012 at 9:59 am

    what is the answer to bridge over troubled students and how do you work it out



  31. Steven Miller on April 11, 2012 at 5:30 am

    email me at sjm1 AT williams.edu — tried emailing y ou but it bounced. //s



  32. James on April 10, 2012 at 9:54 pm

    wow this is a very hard solution… wish i do really know the answer lol this is pretty HARD



  33. Steven Miller on April 2, 2012 at 1:38 pm

    But this takes 20 minutes — there’s a way to do it in 17. Email me at sjm1 AT williams.edu if you want to chat more. //s



  34. E on April 2, 2012 at 6:31 am

    1&4 – 10
    1 – 11
    2&3 – 16
    2 – 18
    1&2 – 20
    Didn’t think I could do that.



  35. Steven Miller on February 11, 2012 at 1:36 am

    email me at [email protected] for a hint ..s



  36. Anonymous on February 10, 2012 at 4:03 pm

    im confused



  37. Steven Miller on February 4, 2012 at 1:57 am

    yes, but there’s a way to do it in just 17 minutes! email me at [email protected] if you want a hint.



  38. awesome on February 3, 2012 at 11:54 pm

    easy, i think
    man one 1 min
    man two 2 min
    man three 5 min
    man four 10 min

    1 and 2 go, 2 min
    1 comes back, 1 min
    1 and 3 go, 5 min
    1 comes back, 1 min
    1 and 4 go, 10 min

    in total=19 minutes!!!



  39. Steven Miller on January 29, 2012 at 2:09 pm

    nope, but your second attempt was correct (not posting as its the soln)



  40. big brain on January 29, 2012 at 9:46 am

    first number 2 and 5 (5min) number 2 come back (7min)he take number 1 (9min)
    number 1 come back (10min) he take number 4 (2omin)

    hahhahaha



  41. Steven Miller on January 12, 2012 at 3:40 pm

    you can do it with binomial coefficients, but this isn’t the right approach (though the answer happens to be close…).



  42. Steven Miller on January 12, 2012 at 2:06 am

    To: Anonymous: Submitted on 2012/01/12 at 2:02 am: Correct, well done! Not posting as it’s the soln.



  43. Anonymous on January 12, 2012 at 2:02 am

    1 and 2 go…1 comes back…5 and 10 go…2 comes back…1 and 2 go again. 17!



  44. Steven Miller on January 11, 2012 at 4:56 pm

    J-crew: correct, well done. not posting as it’s the soln. (You can also email solns to be at [email protected]). //s



  45. Steven Miller on December 19, 2011 at 4:57 am

    unknown (72.45.58.222 Submitted on 2011/12/18 at 2:16 pm): well done, not posting as this is the soln



  46. Steven Miller on December 9, 2011 at 3:34 pm

    You have the wrong values — the third person takes 5 minutes to go over, the fourth takes 10. This gives a total time of 19 minutes I believe.



  47. Ron Freeze on December 9, 2011 at 3:02 pm

    Man 1 and 2 go across = 2minutes
    Man 1 goes back across = 3minutes
    Man 1 and 3 go across = 5minutes
    Man 1 goes back across = 6minutes
    Man 1 and 4 go across = 16minutes
    Everyone is across in 16 minutes



  48. Steven Miller on December 5, 2011 at 2:57 pm

    there’s actually two solutions, but they’re essentially the same



  49. Clay on December 5, 2011 at 2:54 pm

    it took me a while before i got it



  50. Steven Miller on November 11, 2011 at 2:16 am

    email me at [email protected] if you want a hint. ..s



  51. Soroush on November 11, 2011 at 12:57 am

    Oh wait… I just found out how to do it in 18 minutes… but I still can’t figure out how to do it in 17 minutes!



  52. Soroush on November 11, 2011 at 12:55 am

    I found out how to do it in 19 minutes… this one is difficult.



  53. Steven Miller on November 6, 2011 at 3:24 pm

    I’ll email you a hint



  54. jess on November 6, 2011 at 7:08 am

    It takes 20 minutes! Ok….how do youdo this one?



  55. Steven Miller on October 30, 2011 at 12:28 pm

    to Anonymous
    123.238.226.142
    Submitted on 2011/10/30 at 5:03 am
    CORRECT, WELL DONE! Not posting as it’s the soln.



  56. Steven Miller on October 11, 2011 at 1:59 am

    Email me for a hint/soln



  57. Anonymous on October 11, 2011 at 1:02 am

    still can not find the answer



  58. Steven Miller on September 23, 2011 at 1:53 pm

    To Kofi (posted on 2011/09/23 at 12:03 pm): correct!



  59. Kofi on September 23, 2011 at 12:03 pm

    Man 1 and Man 2 go 2mins
    Man 1 returns 1min
    Man 3 and Man 4 go 10min
    Man 2 returns 2mins
    Man 1 and Man 2 go 2mins

    Gives total of 17mins



  60. Steven Miller on July 28, 2011 at 1:02 pm

    Hint: 5 and 10 need to walk together….



  61. malar on July 28, 2011 at 9:41 am

    as of i know, it takes 19 mins to cross… how is it possible within 17 mins



  62. Steven Miller on July 18, 2011 at 1:16 am

    Glad you enjoyed it. //s



  63. Fred on July 15, 2011 at 9:20 am

    that’s so awesome article , thank you 🙂



  64. Steven Miller on March 16, 2011 at 1:27 pm

    Correct answer! You could also switch who comes back first. I normally don’t post answers as then others see the answers, and when answers are online people tend to go to them first without the joy of trying the riddle. You can always send me answers at sjm1 AT williams.edu.



  65. Steven Miller on March 16, 2011 at 1:26 pm

    For which riddle? Sorry for the delay in answering — the system hadn’t told me there was an attempted post! My email is sjm1 AT williams.edu



  66. Steven Miller on March 16, 2011 at 1:26 pm

    Usually not as then people will be able to see the answer, and that removes a lot of the fun.



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