What is big grey and lives in california riddle

A museum has a set of rare chameleons colored blue, red, and green. When two chameleons of different colors meet, they both turn into the color of the other chameleon (if a red and blue met, for example, they would both turn green).

The exhibit currently has 13 blue, 15 red, and 17 green chameleons. The number can change as the chameleons mingle. Can all of the chameleons ever be a single color? If so how can that happen? If not, why is it not possible?

The museum does not add or remove chameleons from the exhibit.

Watch the video for the solution.

Can You Solve The Chameleon Riddle?

Or keep reading below.

"All will be well if you use your mind for your decisions, and mind only your decisions." Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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Answer To The Chameleon Riddle

This is a classic puzzle that relies on the concept of an invariant–a property that does not change even when the parameters do.

For example, the total number of chameleons is 13 + 15 + 17 = 45. While the number of chameleons of each color can change, the total number of chameleons always stays the same at 45. The total number of chameleons is invariant to the color-changing meetings.

This is an obvious point: since the exhibit always has the same number of chameleons, the total number of chameleons never changes. Another way to see this is by mathematical formula. At the start, we have:

blue + red + green = 45

There are 3 possibilities for color-changing. A blue can meet a red, a blue can meet a green, and a red can meet a green. In each case, two of the color types decrease by 1, and the third color type increases by 2. The three possibilities can be represented by:

(blue – 1) + (red – 1) + (green + 2) = 45 (blue – 1) + (red + 2) + (green – 1) = 45

(blue + 2) + (red – 1) + (green – 1) = 45

The decrease in number for the colors that meet is exactly offset by the increase in number for the other color.

Returning to the original problem, after some experimentation, it is evident that the chameleons will not be able to “pair” up correctly so they can all become one color. We need a way to prove this.

To solve the problem, let us find an invariant that considers the pairing of red and blue chameleons. We will consider the simple difference between red and blue chameleons:

red – blue

How does this change as the chameleons mingle?

There are 3 possibilities for color-changing. A blue can meet a red, a blue can meet a green, and a red can meet a green. Let us calculate what happens to the quantity red – blue for those possibilities in order:

(red – 1) – (blue – 1) = red – blue (red + 2) – (blue – 1) = red – blue + 3

(red – 1) – (blue + 2) = red – blue – 3

We have shown the difference can either be the same, or it can go up or down by 3. This makes sense: when a blue and a red meet, the number of each drops by 1 so the difference stays the same. In the other cases, one color drops by 1 and the other color increases by 2, for a net difference of plus or minus 3.

This is the key insight: the difference between red and blue is always the same as the start, plus or minus a multiple of 3.

In our problem we start with 13 blue and 15 red, so:

red – blue = 2

Regardless of how the chameleons meet, this difference will always be 2 plus a multiple of 3. So we have:

red – blue = 2 + 3k, for some integer k

Is it possible the chameleons will ever be the same color? There are 3 ways this could happen:

(45 blue, 0 red, 0 green), so red – blue = -45 = 3(-15) (0 blue, 45 red, 0 green), so red – blue = 45 = 3(15)

(0 blue, 0 red, 45 green), so red – blue = 0 = 3(0)

If the chameleons all became the same color, then the difference red – blue would be a multiple of 3.

But this is not possible in the exhibit! Why? We showed the difference red – blue will never be a multiple of 3–it will always be 2 more than a multiple of 3.

Therefore it is not possible for the 13 blue, 15 red, and 17 green chameleons to ever all become the same color.

There are other ways to present the solution. See more at Cut The Knot: www.cut-the-knot.org/blue/Chameleons.shtml

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