In a board game, we have 11 red, 7 blue, and 20 green tokens. The bank exchanges one red and one blue token for two green tokens, one red and one green token for two blue tokens, and one blue and one green token for two red tokens. During the exchanges, we aim to have all tokens of the same color. Which color will this be?
Solution. Notice that no matter which two colors we change, the sum of the differences in the number of marbles, when divided by 3, always gives the same remainder. For example,

| originally | 11 red |  | 7 blue |  | 20 green | marbles we have, |
| :--- | ---: | :---: | ---: | :---: | :---: | :---: |
| their differences |  | 4 |  | 13 |  | and $20-11=9$, |
| their sum | remainder when divided by 3: | $2 ;$ |  |  |  |  |
| or | 1 | 1 red | and | 1 blue |  | marbles we take away, |
| remaining | 10 red |  | 6 blue |  | and 22 green |  |
| their differences: |  | 4 |  | 16 | and 12, |  |
| the sum's remainder when divided by 3: | 2. |  |  |  |  |  |

This is clearly always true, since if we take away 1 marble of each of two colors, their difference does not change; if, however, we take away 1 marble of one color, the number of marbles of some other color increases by 2, so their difference changes by 3.

We have a total of 38 marbles, and the remainder when divided by 3 is 2. Therefore, the color that cannot remain at the end is the one for which there exists another color such that the difference in the number of marbles is a multiple of 3. The differences in the number of marbles are: $20 g-11 r=9, 20 g-7 b=13, 11 r-7 b=4$. Thus, neither red nor green marbles can remain at the end, only blue.

A set of all blue marbles is indeed achievable in the following way: $11 r, 7 b, 20 g$. We take away 3 blue and 3 green marbles, leaving us with $17 r$, $4 b$, and $17 g$ marbles. Then we exchange $17 r$ and $17 g$ for 34 blue marbles, resulting in 38 blue, 0 green, and 0 red marbles.