Problematics | Walking on an escalator

Suppose you walk up a escalator that is itself moving up. Do you count more steps or fewer compared to moving up a stationary staircase of the same length? What happens if you walk down while the escalator moves up, or vice versa? How does the number of steps you walk change as you vary your walking speed, or the number of steps you cover with each stride?

Puzzles with escalator steps offer a number of variations. As far as I can remember, they have been used only once in Problematics, and that was in the early days of this column in its current form in HT, close to two-and-a-half years ago. So here’s a selection for you, picked up from various published sources and adapted to Problematics-like situations.

#Puzzle 139.1

#139.1A: Professor Pandit walks up a small escalator that is also moving up. Walking one step at a time, she takes 16 steps to reach the top. As an experiment, she walks downstairs on a regular staircase. She then takes the upward-moving escalator again. This time, she walks up two steps at a time, and reaches the top with 12 steps. She takes the staircase down again, but unfortunately cannot continue the experiment because a power failure results in the escalator shutting down.

How many steps does Professor Pandit see on the stationary escalator?

#139.1B: In a mall not far away, Professor Pandit’s colleague Lecturer Lalaji is conducting his own experiment on another escalator, this one moving downward. Like his colleague, Lecturer Lalaji begins by walking in the same direction as the escalator. Walking slowly downward, he reaches the bottom in 50 steps. The next step of his experiment is different, though. From the bottom, he runs up the down-moving escalator. His upward speed now is 5 times his previous downward speed, but he still maintains the discipline of travelling only one step at a time. To accelerate so much without increasing his stride length, indeed, tells us volumes about Lecturer Lalaji’s fitness, even though he is not getting any younger. But to come back to more relevant details, he takes 125 steps to reach the top. He would have loved to continue his experiment, but the manager of the mall has observed him on CCTV. “These professors!” the manager mutters, and switches off the escalator, much to Lecturer Lalaji’s disappointment.

How many steps are visible on the stationary escalator?

#139.1C: In yet another location, two students are conducting an escalator experiment assigned by their teachers Professor Pandit and Lecturer Lalaji. Young Ms Ahmed is walking downward on an upward escalator while her classmate Mr Basu is walking upward on the same escalator. In the time that Mr Basu takes one step, Ms Ahmed takes three. She counts 150 steps from top to bottom, he counts 75 from bottom to top. “How many steps will we see if the escalator’s switched off?” the young man wonders. “That’s easily calculated,” his classmate says.

How many steps?

#Puzzle 139.2

The two professors we met on the above escalators, incidentally, are both family persons and have two children each. Each child is either a boy or a girl, and all four are friends, we are told.

If Professor Pandit’s elder child is known to be a girl, what is the probability that she has two daughters? And if we are told that at least one of Lecturer Lalaji’s children is a boy, what is the probability that he has two sons?

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 138.1

There are several ways that you can avoid relying on longitudes and still explain why any traveller, like Phileas Fogg, will see the sun rise 80 times if they go around the Earth in 79 days, travelling eastward. It was a pleasure going through the various approaches that readers took:

Hi Kabir,

Here is a simple way to explain why a traveller going around the globe in 79 days, travelling eastward, will see the sun rise 80 times. The sun's angular speed with respect to the Earth is 1 rotation per day, westwards, and the traveller's angular speed with respect to the Earth is 1 rotation per 79 days or 1/79 rotations per day, eastwards. Since the two are rotating in opposite directions, their relative angular speed with respect to each other is the sum of their individual speeds = (1 + 1/79) rotations per day = 80/79 rotations per day. Therefore, in the 79 days that the traveller takes for one round, he will see the sun going around 79 x (80/79) = 80 times.

— Professor Anshul Kumar, Delhi

***

One rotation of the Earth involves one sunrise. When a traveller is also moving in the easterly direction, he adds one rotation by coming back to the starting point. Hence one extra sunrise. Anybody travelling east and covering the distance in N days will see N + 1 sunrises.

— Kanwarjit Singh, Chief Commissioner of Income-tax, retired

***

Think of an experiment where you are on a merry-go-round, and your friend is standing outside and watching you. You are running in the same direction as the platform, say clockwise. Now if you move a distance of x and the platform travels a distance of y, the effective distance travelled is (x + y). This gives an analogy of our Problematics puzzle. The traveller and the Earth are rotating in the same direction. So for an outside observer (sun), the total distance covered by the traveller is the sum of the distances by the traveller and the Earth = (1 round by the traveller) + (79 rounds by Earth in 79 days) = 80 rounds in total.

— Sampath Kumar V, Coimbatore

***

Imagine you’re on a 12-hour analog clock, and each full revolution is a day. If you move one hour on the clock face each day, you’ll complete the circle in 12 days, but you'll have seen the hour hand 13 times. Apply the same idea to Fogg's travel.

— Vinod Mahajan, Delhi

[In Vinod Mahajan’s analogy, the Earth is presumed to be stationary while the sun rotates around it. The clock face represents the stationary Earth. The sun (hour hand) and the traveller (you on the clock) must move in opposite directions for this analogy to work. — KF]

#Puzzle 138.2

Hi Kabir,

Other than 3-4-5 (all three sides single digits), the only Pythagoras triple with two integer sides of single digits is 6-8-10. To prove that the solution is unique, we can check all possible combinations with two or all three sides being single digits. This shows that (3, 4, 5) and (6, 8, 10) are unique.

— Shishir Gupta, Indore

To make it clear, a rigorous proof does exist. My sincere thanks to Professor Anshul Kumar and Sampath Kumar V, who have satisfied my curiosity by sending two different proofs to show that the solution is unique. These are, however, too lengthy to present in this column. As Shishir Gupta and a couple of other readers have pointed out, merely examining at the finite number of possibilities is as good a method as any.

Solved both puzzles: Professor Anshul Kumar (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Sampath Kumar V (Coimbatore), Vinod Mahajan (Delhi), Shishir Gupta (Indore), Dr Sunita Gupta (Delhi) Yadvendra Somra (Sonipat), Dr Sunita Gupta (Delhi), Shri Ram Aggarwal (Delhi), Aishwarya Rajarathinam (Coimbatore), Ajay Ashok (Delhi)

Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com.