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Seven Elevators Stop at Six Floors

```Date: 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
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

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
```
Associated Topics:
High School Permutations and Combinations
High School Puzzles

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