Seven Elevators Stop at Six FloorsDate: 09/14/2002 at 12:07:28 From: Jean-Marie Subject: Problem Solving At a party, the superintendent of an apartment building tells you that his building has seven elevators. Each elevator stops on at most 6 floors. He also tells you that if you take the right elevator, you can get to any one floor from any other floor without changing elevators. What is the greatest number of floors that the building could have? I figured out that there are 105 pairs of floors in a 15-story building by taking the number of elevators, seven, and multiplying that by the 15 floors. I also figured that you get the same number if you take the number of floors and multiply it by that same number plus one and divide that by 2. Date: 09/17/2002 at 12:00:29 From: Doctor Wilkinson Subject: Re: Problem Solving Hi, Jean-Marie. You have made some good observations and seem to realize that you can't get more than 15 floors, since one elevator provides access between only 15 pairs of floors, so seven elevators can provide access between only 105 pairs of floors, which is (15*14)/2, as you noted. But 15 is not the answer, because if you add up all the stops on all the elevators, you get 7*6 = 42, and that's not a multiple of 15, but you can see fairly easily that each floor needs to be represented the same number of times. So maybe the answer is 14. This requires working out how to arrange the stops on the various elevators. I worked with this for a while before figuring out a way to do it, and if I hadn't already known something about this kind of problem, I don't know if I could have done it. I'll try to explain it as well as I can. Suppose you had only 3 floors per elevator instead of 6. Then it's not too hard to work out an arrangement that allows you to handle 7 floors: 1 2 3 1 4 5 1 6 7 2 4 6 2 5 7 3 4 7 3 5 6 This is known as "the projective plane of order 2" and it was because I knew about this that I was able to solve the puzzle. Now to get 14 floors with 6 floors per elevator, you just have to tack on another identical arrangement, except with the numbers 8-14 instead of 1-7. That is, just add 7 to the numbers in the first arrangement: 1 2 3 8 9 10 1 4 5 8 11 12 1 6 7 8 13 14 2 4 6 9 11 13 2 5 7 9 12 14 3 4 7 10 11 14 3 5 6 10 12 13 which you can check to see that it solves the problem. - Doctor Wilkinson, The Math Forum http://mathforum.org/dr.math/ Date: 09/17/2002 at 14:13:14 From: Jean-Marie Subject: Thank you (Problem Solving) Dear Dr. Wilkinson, Thank you so much for helping me in solving the problem. Your explanation was great, and easy to understand. I truly appreciate you taking your time to help me. There should be more kind people in the world like you! Thanks again. Sincerely, Jean-Marie |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/