6. Piglet ran down a moving escalator and counted 66 steps. Then he ran up the same escalator at the same speed relative to the escalator and counted 198 steps. How many steps would he have counted if he had gone down a stationary escalator?

Answer. 99.
Let \( u \) be the speed of Piglet, \( v \) be the speed of the escalator (both measured in steps per unit time), and \( L \) be the length of the escalator (in steps). Then, the time Piglet spent descending the moving escalator is \( \frac{L}{u+v} \), and during this time, he counted \( \frac{L}{u+v} \cdot u \) steps. For the ascent against the movement, the time spent was \( \frac{L}{u-v} \), i.e., he counted \( \frac{L}{u-v} \cdot u \) steps. We obtain the equations \( L u = 66 u + 66 v \) and \( L u = 198 u - 198 v \), from which \( 66 u + 66 v = 198 u - 198 v \), hence \( u = 2 v \). Then, \( L u = 66 u + 33 u \), so \( L = 99 \).