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Fifteen white rhinoceroses are standing in a line. Each of them weighs an integer number of pounds. For each rhinoceros (except the leftmost one), its weight plus twice the weight of its left neighbor exactly equals $15000$ pounds.

Determine all possible weight distributions for these fifteen animals!

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    $\begingroup$ Did you mean 15000 rather than 15.000? Using English decimal notation, 15.000 = 15. $\endgroup$ – AndyT Nov 24 '15 at 10:28
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    $\begingroup$ It's 15, because they're standing in a single file line! $\endgroup$ – Samthere Nov 24 '15 at 11:29
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    $\begingroup$ Not sure if you're talking about Northern or Southern Rhinos, but if you're talking about Northern white rhinoceros, you should really probably let someone know you found 15 of them! $\endgroup$ – Timmy Nov 24 '15 at 12:29
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    $\begingroup$ @Timmy I came here hoping someone made a comment like this, but I'm happy to see you put a positive spin on it! $\endgroup$ – Raystafarian Nov 24 '15 at 16:14
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The only possible weight for each rhinoceros is 5000 pounds.

As for why it is the only solution:

First we look at the obvious solution of every rhinoceros weighing the same.
Then the amount they would have to weight would be:
$3w = 15000$
$w = 5000$

Now lets add to that a little difference $d$ for the first weight and this difference will cascade through to the other weights.
$w_0 = 5000 + d_0$

And for all others:
$w_n = w + d_n$

$w_n + 2w_{n-1} = 15000$
$w + d_n + 2(w + d_{n-1})= 15000$
$3w + d_n + 2d_{n-1} = 15000$
$d_n + 2d_{n-1} = 0$
$d_n = -2d_{n-1}$

Therefore:
$d_n = d_0 * -2^n$

Now even for the smallest difference of $d_0 = \pm1$ the 14th and 15th rhinoceros will have a difference of $d_{13} = \pm8192$ and $d_{14} = \pm16384$ respectively.
Since both have an absolute amount greater than $5000$ and at least one of them has to be negative that will also make the respective rhinoceros have a weight of less than zero which should in no way be possible.

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In order for the total weight to be 15 pounds (15.000 = 15, using English decimal notation), with each rhino weighing a positive integer number of pounds the only solution is:

They all weigh 5 pounds.


In order for the total weight to be 15000 pounds (seems more likely if I remember how much a pound is and how big a rhino is), assuming each rhino must have a positive weight, the only solution is:

They all weight 5000 pounds.

This is because:

The "error" (i.e. the difference between the actual weight of a rhino and the perfect weight of 5000) doubles with each rhino while alternating sign. i.e. If we have an "error" of +1 in the weight of the first rhino (i.e. it weighs 4999 or 5001) then the "error" in the second rhino is -2, and in the third is +4 etc. As we have 15 rhinos, if the error of the first rhino is a then the error of the last rhino is 2^14 * a. Hence an "error" of 1 (the smallest possible) in the first rhino gives an error of 8192 in the fourteenth and 16384 in the fifteenth rhino: 5000 minus either of these is negative and hence not possible.

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  • $\begingroup$ But what is about the leftmost one? $\endgroup$ – H.Modh Nov 24 '15 at 10:30
  • $\begingroup$ @H.Modh - What about it? There are no restrictions on it, other than that twice its weight, plus the weight of the one to the right, must equal 15.000. $\endgroup$ – AndyT Nov 24 '15 at 10:32
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Each rhino weighs 5000 pounds. If the leftmost rhino weighs even 5001 or 4999 pounds, the rhinos at the right weigh too much to continue the sequence.

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  • $\begingroup$ @AndyT I agree, those could definitely be space rhinos. $\endgroup$ – dmg Nov 24 '15 at 10:49
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The other answers are right,

All weigh $5$ pounds.

Trial and error is by far the simplest approach. Assume values of the first rhino, and then calculate onwards.

$1000,13000,error$

$2000,11000,error$

$3000,9000,error$

$4000,7000,1000,13000,error$

$5000,5000,5000,\dots$ Solution found!!!

$6000,3000,9000,error$

$7000,1000,13000,error$

$8000,error$

$9000,error$

And so on....

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  • $\begingroup$ That's a pretty bad way of doing your trial and error though. By default, you should start with 3000 (15000/3), and from there check +/- X, reducing X until you reach 1 and see that even with the first Rhino weighing 5001 or 4999, it fails. $\endgroup$ – Julien Blanchard Nov 24 '15 at 18:01
  • $\begingroup$ @JulienBlanchard 15000/3 is what now? $\endgroup$ – dmg Nov 25 '15 at 8:32
  • $\begingroup$ @dmg welp... and It's too late for me to edit. $\endgroup$ – Julien Blanchard Nov 25 '15 at 13:39

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