Fifteen white rhinoceroses

Question

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!

Answer 1

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.

Answer 2

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.

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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.

Answer 3

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.

Answer 4

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....