# Can You Solve The Chameleon Riddle?

A museum has 13 blue, 15 red, and 17 green chameleons. Whenever two chameleons of different colors meet, they both change into the color of the other chameleon (if blue and red met, for example, they would both turn green). As the chameleons mingle, is it ever possible that they all end up being the same color? If so, how? If not, why not? The video presents a solution to this classic puzzle.

Blog post: http://wp.me/p6aMk-4zA

If you like my videos, you can support me at Patreon: http://www.patreon.com/mindyourdecisions

Connect on social media. I update each site when I have a new video or blog post, so you can follow me on whichever method is most convenient for you.

My Blog: http://mindyourdecisions.com/blog/

Twitter: http://twitter.com/preshtalwalkar

Facebook: https://www.facebook.com/pages/Mind-Your-Decisions/168446714965

Google+: https://plus.google.com/108336608566588374147/posts

Pinterest: https://www.pinterest.com/preshtalwalkar/

Tumblr: http://preshtalwalkar.tumblr.com/

Instagram: https://instagram.com/preshtalwalkar/

Patreon: http://www.patreon.com/mindyourdecisions

Newsletter (sent about 2 times a year): http://eepurl.com/KvS0r

My Books

“The Joy of Game Theory” shows how you can use math to out-think your competition. (rated 4/5 stars on 23 reviews) https://www.amazon.com/gp/product/1500497444

“The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias” is a handbook that explains the many ways we are biased about decision-making and offers techniques to make smart decisions. (rated 5/5 stars on 1 review) https://www.amazon.com/gp/product/1523231467/

“Math Puzzles Volume 1” features classic brain teasers and riddles with complete solutions for problems in counting, geometry, probability, and game theory. Volume 1 is rated 4.5/5 stars on 11 reviews. https://www.amazon.com/gp/product/1517421624/

“Math Puzzles Volume 2” is a sequel book with more great problems. https://www.amazon.com/gp/product/1517531624/

“Math Puzzles Volume 3” is the third in the series. https://www.amazon.com/gp/product/1517596351/

“40 Paradoxes in Logic, Probability, and Game Theory” contains thought-provoking and counter-intuitive results. (rated 4.9/5 stars on 7 reviews) https://www.amazon.com/gp/product/1517319307/

“The Best Mental Math Tricks” teaches how you can look like a math genius by solving problems in your head (rated 4.7/5 stars on 3 reviews) https://www.amazon.com/gp/product/150779651X/

“Multiply Numbers By Drawing Lines” This book is a reference guide for my video that has over 1 million views on a geometric method to multiply numbers. (rated 5/5 stars on 1 review) https://www.amazon.com/gp/product/1500866148/

source

An interesting question is:

How many chameleons of each colour must be there, so that you can solve the puzzle? Are there rules?

Requirement is an even number of chameleons in total, but that is not enough.

E.g. red = 10 … blue = 20 …green = 40 DOESN'T WORK

E.g. red = 10 … blue = 20 …green = 50 WORKS

Silly mathematicians, thinking chameleons can never meet in groups larger than 2.

I don't understand this problem…! how come it's impossible to have all in one colour ????

It is possible to make all the chameleons the same color if and only if there is the same number of chameleons for at least two different colors.

Question: what if:

1) 13 blue meet 13 red ->> 0 blue, 2 red, 43 green

2) 1 green meets 1 red ->> 1 blue, 1 red, 42 green

3) blue meets red ->> all are green?

Wouldnt that be the answer to the riddle? 🤔

your wrong

#1 match 15 red and green. 43:0:2

#2 match 1 blue and green. 43:1:1

#3 match 1 red and green. 45:0:0

it is possible

yes very easy 0:56

Any one who isnt able to solve this is stupid af, just pair all the blues with reds, then pair one red with one green and then the new blue one with the remaining red one, all are green… why this long explanation ?????

oh my god, it's not that hard to explain. I was doing this w my boyfriend and we got it in minutes and we just stayed a bit longer bc we looked at the time of the explanation and thought there was more to it, but no. It cant bc there will always be chameleons left and if they meet w another colour they change and that over and over, why did you the the 3k thing?

I went on Paint and drew colored dots to figure this out… Well I guess I'm not very good at math because I did not understood the explications xD

fuck you hurt my brain

What a shitty fckn riddle….i came here for genuine brain wrestling…not bitches cooking up grits….fcin moron…I'm not gonna waste my breath on you…retard!

This was uploaded on my Birthday

look mix 1 blue and 1 green this makes blue have 12 red have 16 and green have 16 then mix all red and green and bam you have 45 blue

this was easy 2 green with 2 red to make 15 blue 13 red and 15 green then pair green and blue

2+2 is also a mutiple of 3. it's 2+2 = 4 + (3×0)

You can do r/g and get 14 b 14 r and 16 g then 14 r and 14 b go together and make 45 green

threesome is the answet

well, the total chameleons is 45, which is an odd number, which means you'll always have 44 same colored chameleons and 1 left over, since it has no one to pair it. WHY THE FUCK WITH THE COMPLICATED ASS EXPLANATIONS

this is boring

Plot Twist the chameleons are dead since museums only keep dead animals

Can't they just change their color to one color……

I figured this out.

A singuler Red and Green Chameleon meet and become Blue.

Now Blue = 14, Red = 14, Green = 16

Then all the Blue and Red Chameleon's meet and become Green leaving you with 45 Green Chameleon's.

There's an odd number of chameleons.

just get the leftover different ones and make them face each other at the same time. for example, 3 reds would react to a blue one at the same time, because none will stand there and wait for the blue one to finish transforming with another one. it's simple.

kill two red chameleons

Am i the only one who read it wrong and thought it said charmeleon in pokemon?

there are odd numbers of total chameleons so there will always be one lonely af chameleon so it's not possible

Hello Presh,

I think there is the way to solve this. If all 13 blue chameleons mingle with the green, 4 green chameleons will be left, 15 + (13 x 2)= 41 red, and 0 blue. If two of the four green chameleons now mingle with 2 red, there will be 2 blue and 39 reds. Then, the remaining two green chameleons mingle with the other 2 blue, and all of them become red! Please write back.

yes but if you have a series of events ?

Manage to have 1,22,22 and at the next event all colors the same…