A man walks down a down-moving escalator and reaches the bottom after taking 50 steps. As an experiment, he then runs up the same escalator, one step at a time, and reaches the top after taking 125 steps. Assuming that he goes up five times as fast as he goes down (that is, he takes five steps going up in the same amount of time as he takes one step going down), and that he makes each trip at a constant speed, how many steps of the escalator are visible if it stops running?
Let n be the number of steps visible when the escalator is not moving, and let a unit of time be the time it takes the man to walk down one step. If he walks down the down-moving escalator in 50 steps, then n – 50 steps have gone out of sight in 50 units of time. It takes him 125 steps to run up the same escalator five times as fast. In this trip, 125 – n steps have gone out of sight in 125 / 5, or 25, units of time. Because the escalator can be presumed to run at a constant speed, we have the following linear equation that readily yields a value for n of 100 steps:
(n – 50) / 50 = (125 – n) / 25
We’d love to hear from you! E-mail us at games@sciam.com to share your experience.
A version of this puzzle originally appeared in the May 1959 issue of Scientific American.