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While being indescribably grateful for the hundreds of Car Talk Puzzlers, I have a quibble with one in particular.   The goal now is to improve, in my opinion, on that puzzle’s official solution.

    Paraphrase of The Dark and Stormy Night from Car Talk Radio

1. All residents of a 3-storey apartment house are home at nightfall as the rain begins.

2. Under a leaking roof, those on the top floor (X many) try to flee to the middle floor, which has only enough room to accept as many neighbors as it has residents (Y many). The rest return upstairs.

3. Soon the middle floor's ceiling begins to leak so everyone there tries to flee to the bottom floor. Again, only as many are admitted as reside there (Z many) while the rest retreat to the middle floor.

The story so far:                           .'  .' .' .' .'.'  .' .'  .' .'.' .' .'  .' .'
        ______________                    .'  .' .' .' .'.'  .' .'  .' .'.' .' .'  .' .' .
       /  \           \                 .'  .' .' .' .'.'  .' .'  .' .'.' .' .'  .' .' .'.
      /    \           \              .'  .' .' .' .'.'  .' .'  .' .'.' .' .'  .' .' .'.'
     /  []  \     X     \           ________________'  .' .'  .' .'.' .' .'  .' .' .'.' .'
    /        \___________\         /  \             \ ' .'  .' .'.' .' .'  .' .' .'.' .'
    | [] []  | []  Y  [] |        /    \             \ '  .' .'.' .' .'  .' .' .'.' .'  .'
    |    _   |           |       /  []  \     X-Y     \  ' .'.' .' __________________ .'.'
    | []| |  | []  Z  [] |      /        \_____________\ .'.' .'  /  \               \ ' .
1.  |___|_|__|___________|      | [] []  | []  Y+Y  [] | .' .' . /    \               \ '
                                |    _   |             |  .' .' /  []  \      X-Y      \ .
                                | []| |  | []    Z  [] | ' .'  /        \_______________\
                          2.    |___|_|__|_____________| .'    | [] []  | []  Y+Y-Z  [] |
                                                               |    _   |               |
                                                               | []| |  | []   Z+Z   [] |
                                                       3.      |___|_|__|_______________|

4. The rain continues until relief arrives to find an equal number of people to rescue on each floor.

Original question.   How many residents — X, Y and Z — began on the top, middle and bottom floors?

Official solution.   (The original puzzle also gives the total number of residents as  X+Y+Z = 12.)

        .' .'.'.' .' .'.' .' .'  .' .' .'  .'.' .'  .'
      .' ______________________'  .' .'  .'.' .'  .'             __________________
    .'  /  \                   \ ' .'  .'.' .'  .'              /  \               \
  .' . /    \                   \ '  .'.' .'  .'               /    \               \
 ' .' /  []  \       X-Y = 4     \ .'.' .   .'                /  []  \     X = 7     \
 .'. /        \___________________\ ' .'  .'                 /        \_______________\
 .'  | [] []  | []  Y+Y-Z = 4  [] | .'  .'                   | [] []  | []  Y = 3  [] |
  .' |    _   |                   |   .'                     |    _   |               |
     | []| |  | []    Z+Z = 4  [] | .'                       | []| |  | []  Z = 2  [] |
 4.  |___|_|__|___________________|'     Official solution.  |___|_|__|_______________|

Looks good, don’t it?   Not in my opinion, so here is a repuzzle whose solution is meant to reveal and redeem my gripe over the original solution.

New question.
Assuming that the solution above is not necessarily the most rational consequence of steps 1, 2 and 3 — and not being given the total number of all residents — what is the smallest possible total and corresponding values for X, Y and Z that would lead to an equal number of rescuees on each floor?

(This is meant as a fairly straightforward situational puzzle rather than a very easy mathematics puzzle.)

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  • $\begingroup$ "Assuming that the official solution above is not the most rational consequence of steps 1, 2 and 3 that would lead to an equal number of rescuees on each floor" - I don't understand what this means. Do steps 1-3 still describe the residents' strategy or not? $\endgroup$ – Rand al'Thor Mar 24 '17 at 22:04
  • $\begingroup$ Define smallest over 3 variables... $\endgroup$ – boboquack Mar 24 '17 at 22:04
  • $\begingroup$ Same strategy, @rand al'thor, thanks, I'll try to restate that more clearly $\endgroup$ – humn Mar 24 '17 at 22:04
  • $\begingroup$ Smallest total, @boboquack, thanks, also to be clarified $\endgroup$ – humn Mar 24 '17 at 22:05
  • $\begingroup$ May we assume X, Y, Z all non-zero and discrete? $\endgroup$ – boboquack Mar 24 '17 at 22:07
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I assume what humn is aiming at is this:

In the original question, after some people have migrated from the middle to the lowest floor, there is now room for more to transfer down there from the top floor. (Is there any reason for them to do so, since the second floor is now getting wet? Well, perhaps the top floor is still worse.)

So

if the floors begin with X,Y,Z inhabitants and have capacities (unknown),2Y,2Z then we should assume that the lower floors end up completely full so that the distribution is X-(Y+Z),2Y,2Z. Now all these are equal so Y=Z and 2Y=X-2Y so X=4Y. Smallest solution is X=4,Y=1,Z=1.

Thus

the sequence of events is as follows. First, one person transfers from top to middle (no more will fit): 3,2,1. Then, one person transfers from middle to bottom (no more will fit): 3,1,2. Now there is space for another in the middle: 2,2,2. And there things stay until the rescue.

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If there's an equal number of residents, then

$X-Y=2Y-Z=2Z$. The second of these equations implies $2Y=3Z$, so we can substitute $Z=\tfrac{2}{3}Y$. Substitute this into the first equation to get $X=3Y-Z=\tfrac{7}{3}Y$.

So the smallest solution is

$X=7,Y=3,Z=2$, the stated solution when the sum is 12,

and any other solution must be a multiple of these values, e.g.

$X=14,Y=6,Z=4$ or $X=21,Y=9,Z=6$.

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  • $\begingroup$ This generalizes the original solution alright but doesn't improve on its official solution $\endgroup$ – humn Mar 24 '17 at 22:10
  • $\begingroup$ @humn I don't understand what you're looking for then. $\endgroup$ – Rand al'Thor Mar 24 '17 at 22:10
  • $\begingroup$ I think there is a solution that is at least as reasonable but also more realistic, without going out of bounds. $\endgroup$ – humn Mar 24 '17 at 22:14
  • $\begingroup$ @humn How? The maths only supports this one solution and its multiples. What do you mean by "going out of bounds" (or for that matter "reasonable" or "realistic")? $\endgroup$ – Rand al'Thor Mar 24 '17 at 22:16
  • $\begingroup$ It's not a strict math puzzle. The best tag I could find was situation $\endgroup$ – humn Mar 24 '17 at 22:17

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