We are restarting the elevator of problem F. 2280 in the ten-story building. It starts again from the ground floor, and during its journey, it will stop exactly once on every floor, just as before. However, this time we also ensure that after stopping at an even-numbered floor, it stops at an odd-numbered floor, and after an odd-numbered floor, it stops at an even-numbered floor. (We consider the ground floor as the 0th floor.) How many meters can it travel at most under these conditions, if the height difference between two adjacent floors is 4 meters?
I. solution. In the solution to problem F. 2280 (Volume 62, No. 4, p. 155, 1981), we saw that if we do not assume that the elevator stops alternately at even and odd numbered floors, it can travel a maximum of 55 floors. If we now also assume that it stops alternately at even and odd numbered floors, then between any two consecutive stops, it travels an odd number of floors. It stops a total of ten times after starting, so the length of the traveled distance (expressed in floors) will be the sum of ten odd numbers, which is even. Therefore, the elevator can travel an even number of floors, at most 54 floors, which is $54 \cdot 4 = 216$ meters. It can indeed travel this much if its consecutive stops are on the 0th, 7th, 2nd, 9th, 4th, 5th, 10th, 3rd, 8th, 1st, and 6th floors.

Gábor Megyesi (Szeged, Ságvári E. Gymnasium, 1st year)

II. solution. In the solution to problem F. 2280, we saw that the elevator can only travel 55 floors if it changes direction after every stop. If we now also assume that it stops alternately at even and odd numbered floors, then these two conditions together mean that the elevator always starts upwards from even numbered floors and arrives from above to even numbered floors, since it starts from the 0th floor. In this case, however, the 10th floor would have to be left out of its route, as it cannot arrive from above. Therefore, it can travel at most 54 floors, and this is achievable if the elevator stops at the 0th, 9th, 2nd, 7th, 4th, 5th, 10th, 1st, 8th, 3rd, and 6th floors.

Jenő Töröcsik (Budapest, Fazekas M. Gymnasium, 3rd year)

Remarks. 1. If we direct the elevator to the farthest possible floor after every stop (and start) under the given conditions, then the first four stops will necessarily be the 0th, 9th, 2nd, and 7th floors. The fifth stop can be either the 4th or the 10th floor, and accordingly, the sequence of subsequent stops will be either 4th, 1st, 10th, 3rd, 8th, 5th, 6th floor or 10th, 1st, 8th, 3rd, 6th, 5th floor. In the first case, the elevator travels only 52 floors, and in the latter case, only 50 floors. This example also shows that in a complex problem, the best result is not necessarily achieved by always choosing the best option at every step.

2. The remark to problem F. 2280 shows that in a $2n$-story building, the elevator can travel a maximum of $n(2n+1)$ floors. If it stops alternately at even and odd numbered floors, then according to the reasoning of the second solution, it can travel less than this, at most $n(2n+1)-1$ floors. If $n$ is odd, this is achievable, for example, with the following sequence: $0., (2n-1)., 2., (2n-3)., \ldots, (n-1)., n., 2n., 1., (2n-2)., 3., \ldots, (n+3)., (n-2)., (n+1).$ floors. If $n$ is even, then $n(2n+1)-1$ is odd, but according to the reasoning of the first solution, the number of floors is even, so in this case, the elevator can travel at most $n(2n+1)-2$ floors. This is achievable, for example, with the following sequence: 0., $(2n-1)., 2., (2n-3)., \ldots, (n-2)., (n+1)., 2n., 1., (2n-2)., 3., \ldots, (n+2)., (n-1)., n.$ floors.