A long time ago, there existed Martians with $3$ different colours: red, green and blue. As Mars was devastated by an intergalactic war, only $2$ Martians of each colours survived. In order to reconstruct the Martian population, they decided to use a machine that transforms two Martians of distinct colours into four Martians of colour different to the two initial ones. For example, if a red Martian and a blue Martian use the machine, they'll be transformed into four green Martians. a) Is it possible that, after using that machine finitely many times, we have $2022$ red Martians, $2022$ green Martians and $2022$ blue Martians? b) Is it possible that, after using that machine finitely many times, we have $2021$ red Martians, $2022$ green Martians and $2023$ blue Martians?
Problem
Source: Lusophon Mathematical Olympiad 2023 Problem 1
Tags: combinatorics, number theory
lpieleanu
28.05.2023 01:02
Note that it suffices to prove there is an algorithm that will transform $2k$ Martians of each color into $2k+2$ Martians of each color for any positive integer $k$.
If we input a red Martian and a green Martian into the machine, we will end up with $2k-1$ red Martians, $2k-1$ green Martians, and $2k+4$ blue Martians.
Now, if we input a green Martian and a blue Martian into the machine (note that we can do this because every color still has at least one Martian), we will end up with $2k+3$ red Martians, $2k-2$ green Martians, and $2k+3$ blue Martians.
Finally, inputting a red Martian and a blue Martian into the machine we will end up with $2k+2$ Martians of each color, so we are done. $\square$
americancheeseburger4281
28.05.2023 03:57
Solution for b) analyzing each operation individually it can be noticed that the difference between each quantity of martians is preserved modulo 5, fact that does not correspond in (2021,2022,2023