There are $2008$ blue, $2009$ red and $2010$ yellow chips on a table. At each step, one chooses two chips of different colors, and recolor both of them using the third color. Can all the chips be of the same color after some steps? Prove your answer.
Problem
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Tags: combinatorics, 2010, PMO
Irrational_phi
16.11.2015 11:06
Try considering the numbers$\pmod{3}$!
First, we consider the numbers modulo $3$. Let there be $B$ blue chips, $R$ red chips, and $Y$ yellow chips. We have that $B \equiv 1 \pmod{3}, R \equiv 2 \pmod{3}, \text{and } Y \equiv 0 \pmod{3}$. Notice that if we take two chips of different colors, and add recolor them to color we haven't picked, all the chips will still have different remainders when divided by 3. For example, if I take 1 blue chip and 1 red chip, recolor them with yellow, we subtract 1 from $R$, subtract 1 from $B$, and add 2 to $Y$. Since $2 \equiv -1 \pmod{3}$, we see that their remainders when divided by 3 all all just subtracted by 1. And so it is impossible for all chips to have the same color!
Takeya.O
02.11.2016 03:46
Let blue point be $0,$red point be $1,$yellow point be $2$.It is easy to show that the total point is invariant in $\pmod{3}$ after each step.At initial state,the total point is $0\cdot 2008+1\cdot 2009+2\cdot 2010\equiv 2\pmod{3}$ If all colors are the same, then the total point is $k(2008+2009+2010)\equiv 0\pmod{3}$ Thus there is no possibility that the all colors are the same.
arandomperson123
03.11.2016 14:47
this is actually a well known combo problem that can be generalized to any $n, n+1, n+2$, and the solution is still looking at modulus 3...