A Slightly Higher Level Problem From Combinatorics

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A Slightly Higher Level Problem From Combinatorics

image containing math problem statement

A problem from combinatorics, slightly twisted but not very difficult.

Problem contributed by debosmita1729

Let $b$ , $g$ , and $r$ denote the number of Blue, Green, and Red chameleons respectively. The total number of chameleons is $N = b + g + r = 13 + 15 + 17 = 45$ .

Consider the meeting of a Blue and a Green chameleon. The population changes as follows:

(b, g, r) \rightarrow (b-1, g-1, r+2)

We look for an invariant property modulo $3$. Let us examine the difference between the counts of any two colors, for instance, $b - g$.

After the transformation, the new difference $(b' - g')$ is:

b' - g' = (b-1) - (g-1) = b - g

The difference remains exactly the same.

Now consider the difference between Blue and Red, $b - r$. The new difference $(b' - r')$ is:

b' - r' = (b-1) - (r+2) = b - r - 3

b' - r' \equiv b - r \pmod{3}

Thus, the difference between the number of chameleons of any two colors is invariant modulo $3$ .

Initial State:

b = 13, \quad g = 15, \quad r = 17

g - b = 15 - 13 = 2 \equiv 2 \pmod{3}

r - g = 17 - 15 = 2 \equiv 2 \pmod{3}

r - b = 17 - 13 = 4 \equiv 1 \pmod{3}

If all chameleons become a single color (e.g., all Red), the state would be $(0, 0, 45)$. The difference between the zero-count colors would be:

0 - 0 = 0 \equiv 0 \pmod{3}

Since we start with differences congruent to $1$ or $2 \pmod{3}$, and the operation preserves these differences modulo $3$, we can never reach a state where the difference is $0 \pmod{3}$. Therefore, it is impossible for all chameleons to become the same color.