# Escalator puzzle equation

I'm trying to understand the escalator puzzle.

A man visits a shopping mall almost every day and he walks up an up-going escalator that connects the ground and the first floor. If he walks up the escalator step by step it takes him 16 steps to reach the first floor. One day he doubles his stride length (walks up climbing two steps at a time) and it takes him 12 steps to reach the first floor. If the escalator stood still, how many steps would there be on sight?

The solution, apparently, is as follows:

$16x = 12(x+1)$, so $x=3$, so the answer is 48.

But why can we say $12(x+1)$?

First, he covers 16 steps and the motion of the escalator gives him a multiplier of $x$ to cover a total of $16x$ steps. That makes sense.

But why is this the same as 12 steps with a multiplier of $(x+1)$?

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There is one glaringly missing piece of information here: Can we assume that he takes the same time per double step as he previously did per single step? The problem text doesn't say, although the problem is unsolvable without this information. Oh well, better assume the answer is yes … –  Harald Hanche-Olsen Dec 20 '12 at 16:27
I'm glad I was told that there are those rare days when the man doesn't visit the mall. /sarcasm –  Greg Martin Mar 31 '13 at 19:31

The quantity $x$ represents how the total distance (measured in escalator-steps) he will travel over the time period it takes him to take one step. When he takes his first step on the escalator, he moves up one escalator-step, plus however far the escalator has moved him in that time. Thus, we could write $x=1+y$ where $y$ is the distance (measured in escalator-steps) the escalator moves him the time period it takes him to take one step.

When he decides to take two steps at a time, in each human-step-time-period he now moves up $2$ escalator-steps under his own power (it doesn't say that explicitly, but I think we are supposed to assume it), plus the aid of $y$ escalator-steps provided by the escalator. Thus, if it now takes him $12$ human-step-time-periods to reach the top, we know that the total distance he traveled is $12(2+y)=12(1+x)$.

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Welcome back!${}$ –  user17762 Dec 20 '12 at 20:47
Thanks, Marvis :) –  Zev Chonoles Dec 20 '12 at 22:06

$x$ is the amount of progress the man makes in the time it takes him to take one step walking regularly. Clearly $x>1$, since he gets one step from his own effort; he then gets an extra $x-1$ steps worth of progress from the movement of the escalator.

When he's moving twice as fast, he still gets $x-1$ steps worth of progress from the escalator for each step he takes, but he now gets two steps worth from his own effort. So the total progress in the time it takes him to take one step is now $(x-1)+2 = x+1$.

When he's walking at regular speed, he takes 16 steps, making $x$ progress each time. When he's walking at double speed he takes 12 steps making $x+1$ progress each time. The end result is the same in the two cases, so we have $16x = 12(x+1).$

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It can also be interpreted by means of Lowest Common multiple of the number of steps in he given two cases. The answer is LCM of $16$ and $12$, which is $48$.

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Let $d$ be the distance traveled, which remains same in both the cases. if $v$ is the speed of the man and $x$ is the speed of elevator, in case 1 the number of steps taken is $$\frac d{v+x}=16$$ In case 2 it is $$\frac d{2v+x}=12$$ because now he is traveling at double the speed; eliminating $d$, we get $x=2v$; therefore $d=48v$; when stationery $x=0$, we get no. of steps as $48v/v=48$.

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