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chameleons

the answer is simple
the equation should be
2X + Y = 45
Where Y is odd. If we can achieve it then problem is solved.
Look at number now:
13 red chameleons
15 green chameleons
17 blue chameleons

Every time they meet (opposite color) they will end up either all even or all odd. say red and green meets
12 red chameleons
14 green chameleons
18 blue chameleons

The above equation can never be true i.e. 2X + Y = 45
where we need two even and one odd.
So it is not possible to achieve it.

Ratan
Monday, April 26, 2004

Original quiz it at location:
http://www.techinterview.org/Puzzles/chameleons.html

Ratan
Monday, April 26, 2004

It is possible

13 red chameleons
15 green chameleons
17 blue chameleons

Assume that a green chameleon meets a blue chameleon and changes color to a red chameleon. So after this, the numbers are changed as

14 red
14 green
16 blue

After this 14 red chameleos meets 14 green ones individually and eventually you have 30 blue chamelons

Mahesh
Tuesday, April 27, 2004

Just to clarify, two chameleons do not meld into one chameleon.  All they do is change color.  So the number of chameleons would stay the same, hence forty five chameleons would not end up to be thirty chameleons.

In the first post, if a red chameleon meets a green chameleon, you would end up with two more blue chameleons.  The numbers should look like this:

13 red chameleons
15 green chameleons
17 blue chameleons

red meets green

12 red chameleons
14 green chameleons
19 blue chameleons

JHY
Tuesday, April 27, 2004

Enumerate the colors -- say red=0, green=1, blue=2. Construct the sum:S= 0*Nred + 1*Ngreen + 2*Nblue. S%3(mod3) is a constant, since it remains the same for each pair:
(0+1)%3 = (2+2)%3,
(0+2)%3 = (1+1)%3,
(2+1)%3 = (0+0)%3.

Initially S=0*14+1*15+2*16, and S%3=2, while for 45 of the same color S%3 would be 0. Done.

DK
Saturday, May 01, 2004

It's not possible, because the distribution of the chameleons is such that they are separated by two.  Every time two different colored chameleons change colors, one of the colors goes up by 2, and the other two colors go down one each.  Because of this, no two colors will ever have the same number of chameleons at the same time.  The numbers will just keep dancing around each other.

eliza s
Thursday, June 10, 2004

mahesh u great, so is the chameleons

deepa v
Thursday, July 01, 2004

My instinct is exactly the same as Mahesh.  There must be something wrong here, otherwise it's too easy for adult.

Steve Li
Thursday, July 01, 2004

Two chameleons dont meld into one. duh

Mohammed Saiyad
Wednesday, July 14, 2004

Initial difference between green and red hameleons
is  Diff(green,red)=numOfGreen-numOfRed=-2

It's obvious, that after each meeting of hameleons, the value of Diff(green,red) - remains unchanged or Diff+=3 or Dif-=3.

In that case, it's obvious too, that the value of Diff(green,red) can't become zero.

Sergey N. Sokolov
Thursday, July 15, 2004

Never Converges..... Although, not acceptable by true mathematical standards... here's my proof by example ;-)

Consider a scaled down version of the problem:

1 Red (R), 3 Green  (G), 5 Blue (B)

R  G  B
1  3  5
0  2  7
--------
4  0  5
0  8  1
2  7  0
--------
0  5  4
8  1  0
7  0  2
--------
5  4  0
1  0  8
0  2  7
--------

Notice how the same numbers keep getting exchanged among the different colors. The dottec lines are to show sets of repeating patterns.

Awaiting comments :-)

EE/ CS Guy
Friday, July 23, 2004

It is obviously capable of happening.  Here is step by step proof:
It is easily possible.

Score: 13 Red Chameleons, 15 Green Chameleons, 17 Blue Chameleons

1 Red Chameleons and 1 Green Chameleons meet

Score: 12 Red Chameleons, 14 Green Chameleons, 19 Blue Chameleons

1 Red Chameleons and 1 Green Chameleons meet

Score: 11 Red Chameleons, 13 Green Chameleons, 21 Blue Chameleons

1 Red Chameleons and 1 Green Chameleons meet

Score: 10 Red Chameleons, 12 Green Chameleons, 23 Blue Chameleons

1 Red Chameleons and 1 Green Chameleons meet

Score: 9 Red Chameleons, 11 Green Chameleons, 25 Blue Chameleons

1 Red Chameleons and 1 Green Chameleons meet

Score: 8 Red Chameleons, 10 Green Chameleons, 27 Blue Chameleons

1 Red Chameleons and 1 Green Chameleons meet

Score: 7 Red Chameleons, 9 Green Chameleons, 29 Blue Chameleons

1 Red Chameleons and 1 Green Chameleons meet

Score: 6 Red Chameleons, 8 Green Chameleons, 31 Blue Chameleons

1 Red Chameleons and 1 Green Chameleons meet

Score: 5 Red Chameleons, 7 Green Chameleons, 33 Blue Chameleons

1 Red Chameleons and 1 Green Chameleons meet

Score: 4 Red Chameleons, 6 Green Chameleons, 35 Blue Chameleons
1 Red Chameleons and 1 Green Chameleons meet

Score: 3 Red Chameleons, 5 Green Chameleons, 37 Blue Chameleons
1 Red Chameleons and 1 Green Chameleons meet

Score: 2 Red Chameleons, 4 Green Chameleons, 39 Blue Chameleons
1 Red Chameleons and 1 Green Chameleons meet

Score: 1 Red Chameleons, 3 Green Chameleons, 41 Blue Chameleons

1 Red Chameleons and 1 Green Chameleons meet

Score: 0 Red Chameleons, 2 Green Chameleons, 43 Blue Chameleons

1 Green Chameleons and 1 Blue Chameleons meet

Score: 1 Red Chameleons, 1 Green Chameleons, 43 Blue Chameleons

1 Red Chameleons and 1 Green Chameleons meet

Score: 0 Red Chameleons, 0 Green Chameleons, 45 Blue Chameleons

Red Chameleon
Wednesday, September 01, 2004

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