The Broken Down Elevator

Observe the following cycle of elevator stops: $1 \to 8 \to 4 \to 11 \to 7 \to 3 \to 10 \to 6 \to 2 \to 9 \to 5 \to 1$ . This guarantees that all $11$ floors are reachable.

To see that $10$ or less floors is not enough, simply look at floor $4$ (or the highest floor if there are less than four floors). An elevator from the fourth floor cannot go down (to floor $0$ ?), nor go up (to floor $11$ ?), so no elevator can go from floor $4$ .

Thus $\boxed{11}$ floors is the minimum.

Here's a general proof that if the up elevator goes up $a$ floors and the down elevator go down $b$ floors then the minimum building has height is $a+b$ .

W.L.O.G. we can assume $a>b$ if not then we can adopt a similar strategy starting from the top floor.

First ride the up elevator to floor $A_1=1+a$ and then ride the down elevator as many times as possible to floor $B_1$ where $1\leq B_1\leq b$ .

Then ride the up elevator once to floor $A_2=B_1+a$ and then ride the down elevator as many times as possible down to floor $B_2$ with $1\leq B_2\leq b$ . Continuing this process $b$ times generates a sequence of numbers $B_1,B_2,B_3,\dots B_b$ with $0<B_i\leq b$ in each case.

We have that $B_i \equiv 1+ka \mod b$ and since $a$ and $b$ are coprime we can show that all of the $B_i$ s are different, and are hence some ordering of the set $\{1,2,\dots b\}$ . The corresponding sequence of $A_k$ s is the some ordering of the set $1+a,2+a, 3+a,\dots b+a\}$ and again since $a$ and $b$ are coprime these numbers are a complete set of residues modulo $b$ .

We can show that each floor, $k$ with $1\leq k\leq a+b$ is visited during this cycle of moves, since each number in this range will be congruent modulo $b$ to one of the $A_k$ s and hence visited on the downward journey from that floor. Since we have that $B_b=1$ , we have found a cycle which takes us to each of the floors, and we are done.

A building of height $a+b-1$ is not possible since there are no moves from floor $b$ .

The problem is with 1st, 2nd, 3rd and 4th floors as the elevator can't go below them. Thus we must find (minimum) 7 floors above them so the elevator can find a path to move. if we find 7 floors above the 4th floor then absolutely we will find 7 floors above 1st, 2nd and 3rd floors. Then the minimum number of floors is $4 + 7 = 11$ floor .

Suppose there are 11 floors. One tactic you could use to access all floors is to start at the top (as the start point isn't specified), go down by 4 twice then up by 7. However, if we have 10 floors, then the 4th floor up is unreachable because there are no elevators that are 7 floors above or 4 floors below.

$\therefore$ ${Minimum}$ = 11