# Fifteen white rhinoceroses

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!

• Did you mean 15000 rather than 15.000? Using English decimal notation, 15.000 = 15. Nov 24, 2015 at 10:28
• It's 15, because they're standing in a single file line! Nov 24, 2015 at 11:29
• 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! Nov 24, 2015 at 12:29
• @Timmy I came here hoping someone made a comment like this, but I'm happy to see you put a positive spin on it! Nov 24, 2015 at 16:14

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.

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.

• But what is about the leftmost one? Nov 24, 2015 at 10:30
• @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. Nov 24, 2015 at 10:32

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.

• @AndyT I agree, those could definitely be space rhinos.
– dmg
Nov 24, 2015 at 10:49

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....

• 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. Nov 24, 2015 at 18:01
• @JulienBlanchard 15000/3 is what now?
– dmg
Nov 25, 2015 at 8:32
• @dmg welp... and It's too late for me to edit. Nov 25, 2015 at 13:39