Problem: A circle is divided in 6 sectors by 3 diameters. Each sector contains a pawn. We are allowed to chose two pawns and move each of them to a sector bordering the one it stands on at the moment. Is it possible to gather all 6 pawns in one sector using such operations?
My solution: Take an arbitrary sector and note that there are always 2 pawns 1 step away from the sector, 2 pawns 2 steps away from the sector and 1 pawn 3 steps away from the sector. In total we have at to perform 9 steps in total to get to the sector. With the given operation the following things can happen:
- -2 steps (we chose 2 pawns and both move them one step closer to the wanted sector)
- +2 steps (we chose 2 pawns and both move them one step away from the wanted sector)
- -1 + 1 v 1-1 =0 steps (we chose 2 pawns, we move one closer to the given sector and 1 one step away from the given sector)
Now note that we always have that the total amount of steps that have to be taken to get all the pawns to the sector mod 2 is 1 mod 2, this is an invariant as with any given operation it stays the same. Suppose that we can put all 6 in one sector, then the amount of steps mod 2 is 0 mod 2. Since this is different from the invariant this can't happen
Is my solution correct? Thanks in advance