No, they can never all be the same colour. Let R, G, and B represent the number of chameleons with each colour. Consider the differences R-G, R-B, and G-B (currently -2, -4, and -2 respectively).
If two chameleons change to the third colour, the numbers R, G, and B will change in some combination of -1, -1, and +2. Each difference will change by (-1) - (-1) = 0, (-1) - (+2) = -3 or (+2) - (-1) = +3.
If x and y are both 0, the difference between them must also be 0, so getting the difference to 0 is a necessary condition. (It is also a sufficient condition, since "n" green and "n" blue chameleons can just pair off and turn red.)
What repeated combination of 0 -3 and +3 will turn -2 or -4 into zero? There is no such combination. So there can never be the same number of chameleons with two different colours. So, there can never be zero chameleons of two different colours.
Thursday, July 25, 2002
I am not sure about the answer to this puzzle . Steve says that no combination of -3,+3 and 0 will lead -2,-2,-4 into 0 .
Well what about -3*-2 + 3*-2 + 0*-4 . Now that is zero .
So i am a little unsure about this reasoning of this puzzle.
Could somebody please make the reasoning more clear ......
Friday, August 23, 2002
Consider the difference between the number of two different colours. This value starts out equal to either -2 or -4. You want to make this value 0. Each step, you can make a move that either leaves the value the same, increases it by 3, or decreases it by 3.
There is no combination of such moves that will give you a result of 0.
Friday, August 30, 2002
If the rules of the combination is as follows:
If the numbers of the chameleons are odd, one of them should sit out the meetings at each step, since only 2 can meet.
Then, I think, that if the number of chameleons is either a power of 2, or exceeds a power of 2 by 1, there is a chance that at some point of time, all of them have the same color.
Since 45 is a number that does not satisfy the above constraint, all the chameleons cannot be of the same color at any point in time.
Please comment on the above.
Saturday, August 31, 2002
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