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I am trying to understand the solution to the below puzzle

There are 13 Red, 15 Green, and 17 Blue Chameleons at some point of

time. Whenever two Chameleons of the different colors meet both of

them change their color to the third color. Is it ever possible for

all Chameleons to become of the same color?

I am able to figure out “manually step by step” that it is not possible to get all in one color.

However I don’t understand the more formal solution using the modulo, for example Solution 1 here.

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I realize that after every change the differennce between the second and first terms is -3, 0 or 3.

What are the vectors (2,2,2) representing?

I don’t understand the final statement there:

`We see that, with the initial vector (1, 0, 2) modulo 3, the only possible`

`combinations of colors on the island are (1, 0, 2), (0, 2, 1), and (2, 1, 0),`

`and these occur cyclically with period 3.`

Can anybody explain in simple way this solution to the problem?

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- How can I quickly know the rank of this / any other matrix?
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The trick is to show that:

$$(17,15,13)+a(-1,-1,2)+b(-1,2,-1)+c(2,-1,-1)$$

is never $(45,0,0), (0,45,0),$ or $(0,0,45)$ for $a,b,c$ integers.

But since $(2,-1,-1)=(-1)(-1,2,-1)+(-1)(2,-1,-1)$, you can eliminate the third term and ask to find $a,b,$ so that we only need to show that:

$$(17,15,13)+a(-1,-1,2)+b(-1,2,-1)$$

can never be $(45,0,0), (0,45,0),$ or $(0,0,45)$ when $a,b$ are integers.

If we have such $a,b$, then $a+b\equiv 17\pmod {45}, 2b-a\equiv -15\pmod{45}$ and thus $3b\equiv 2\pmod{45}$, which is impossible.

Another way to state this is:

If $(x,y,z)+a(-1,-1,2)+b(-1,2,-1)+c(2,-1,-1) = (x’,y’,z’)$ then $x-y\equiv x’-y’\pmod{3}$ and $y-z\equiv y’-z’\pmod{3}$.

So if $(x’,y’,z’)=(45,0,0),(0,45,0)$ or $(0.0,45)$ then $x-y\equiv 0\pmod 3$. But $17-15\equiv 2\pmod 3$.

This also means you can’t get to $(15,15,15)$ since again it would mean $x-y\equiv 0\pmod 3$.

You can see that

Red chameleons mod 3 = 13 mod 3 = 1

Green chameleons mod 3 = 15 mod 3 = 0

Blue chameleons mod 3 = 17 mod 3 = 2

Then, you take all possible cases

If a red chameleon meets a green chameleon, it becomes (0 2 1), which is subtract 1 modulo 3 from the first two components and add 2 modulo 3 on the third component.

If a red chameleon meets a blue chameleon …

If a green chameleon meets a blue chameleon…

Now we suppose that we are in the state (0, 2, 1)

If a red chameleon meets a green chameleon…

and so on

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