The building manager of an Amazon warehouse says that that building has
seven elevators and that each elevator can stop and open at only six floors (does
not have the buttons to stop at the remaining floors). She also says that in that
building you can go from any floor to any other floor by taking only one elevator
(provided you take the correct one) so that the people who work in the
warehouse do not waste time taking multiple elevators.
What is the largest number of floors that such a building can have?




This problem essentially involves finding the maximum number of floors that can be connected by a set of elevators, each of which can stop at only six distinct floors, under the condition that from any floor to any other, you can take exactly one elevator.

We can model this situation mathematically. Let:
- \( F \) be the total number of floors,
- Each elevator stops at 6 floors, so it connects 6 out of \( F \) floors,
- There are 7 elevators in total.

To ensure that any floor can be reached from any other floor via exactly one elevator, every pair of floors must share at least one common elevator. Therefore, for any pair of floors, there must be at least one elevator that stops at both floors.

This setup is very similar to a problem in combinatorics related to **set covers**, where we need to ensure that every pair of elements is covered by at least one set. More formally, we need to distribute the \( F \) floors into subsets (elevators) of size 6 in such a way that each pair of floors appears in at least one of these subsets.

### Calculation:
We can make an estimate based on a combinatorial approach, where every pair of floors (from a total of \( F \) floors) needs to be covered by at least one of the 7 elevators.

The total number of floor pairs is given by \( \binom{F}{2} \), which represents the number of unique floor-to-floor connections. Each elevator covers a subset of \( \binom{6}{2} = 15 \) pairs of floors because it stops at 6 floors, and each pair of floors covered by that elevator forms a connection.

Thus, the total number of floor-to-floor connections covered by the 7 elevators is:
\[
7 \times 15 = 105
\]
We want to cover all possible floor pairs, so we need:
\[
\binom{F}{2} \leq 105
\]
Expanding \( \binom{F}{2} \), we have:
\[
\frac{F(F-1)}{2} \leq 105
\]
Multiplying both sides by 2:
\[
F(F-1) \leq 210
\]
Solving this inequality:
\[
F^2 - F - 210 = 0
\]
Using the quadratic formula:
\[
F = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-210)}}{2(1)}
\]
\[
F = \frac{1 \pm \sqrt{1 + 840}}{2}
\]
\[
F = \frac{1 \pm \sqrt{841}}{2}
\]
\[
F = \frac{1 \pm 29}{2}
\]
So, \( F = \frac{1 + 29}{2} = 15 \) (we discard the negative root).

### Conclusion:
The largest number of floors that such a building can have is **15 floors**.

Would you like further details or have any questions?

Here are five related questions you might consider:
1. How would the number of floors change if there were 8 elevators?
2. What would happen if each elevator could stop at 7 floors instead of 6?
3. Can we use this approach for a building where elevators stop at random floors?
4. What would be the effect if some elevators could skip floors?
5. How can we ensure optimal travel efficiency in buildings with more floors and fewer elevators?

**Tip:** The concept of combinatorial designs, such as block designs, can often provide solutions to problems involving efficient coverage of connections or interactions like in this elevator problem.

The building manager of an Amazon warehouse says that that building has seven elevators and that each elevator can stop and open at only six floors (does not have the buttons to stop at the remaining floors). She also says that in that building you can go from any floor to any other floor by taking only one elevator (provided you take the correct one) so that the people who work in the warehouse do not waste time taking multiple elevators. What is the largest number of floors that such a building can have?

Explore the combinatorial problem of determining the maximum number of floors in a warehouse with 7 elevators, each stopping at only 6 floors, ensuring one-elevator travel between any two floors. This solution utilizes set theory, quadratic equations, and combinatorics. Suitable for high school students in Grades 10-12.

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