Abelkonkurransen 2025 1a

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Abelkonkurransen 2025 1a

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: abelkonkurransen

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

50 posts

#1 h Mar 20, 2025, 11:03 AM • 2 Y

Y by Clearlove7, cubres

Peer and Solveig are playing a game with $n$ coins, all of which show $M$ on one side and $K$ on the opposite side. The coins are laid out in a row on the table. Peer and Solveig alternate taking turns. On his turn, Peer may turn over one or more coins, so long as he does not turn over two adjacent coins. On her turn, Solveig picks precisely two adjacent coins and turns them over. When the game begins, all the coins are showing $M$ . Peer plays first, and he wins if all the coins show $K$ simultaneously at any time. Find all $n\geqslant 2$ for which Solveig can keep Peer from winning.

This post has been edited 3 times. Last edited by Lil_flip38, Mar 20, 2025, 11:05 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1492 posts

#3 h Apr 12, 2025, 8:15 PM • 1 Y

Y by cubres

We claim that that Solvieg can stop Peer for all $n \ge 5$ .
First for $n=3$ just let Peer choose the 1st and 3rd coin on the row and flip them in which case no matter what Solveig does, Peer's next move is just filling the empty.
For $n=4$ its kind of the same thing, let Peer chose 1st and 4th cell and Solveig can either self-lose or let Peer finish depending on how she wants to lose here.
For $n \ge 5$ we are gonna show Solveig's strat which pretty much just follows the principle of trying to not lose on the first move. It is sufficient to show Solveig can stop Peer on $n=5$ since then you just repeat the strategy for all further $n$ 's. so now we do this:
If on first move or later on Peer leaves Solveig a position with one single coin $K$ then Solveig will just take this coin and one $M$ of either side avaliable and toss these two to flip them to keep the number of coins $K$ at one, clearly due to Pigeonhole there is gonna always be two $M$ coins together so Solveig does not inmediately lose.
If on first move or later on Peer leaves Solveig a position with 2 $K$ coins, we will evaluate this depending on their placement, at the beggining there is only $6$ positions Peer can make so that Solveig has to make her move on it, and on the rest we have two adjacent $K$ coins in which case Solveig just flips them both and obviously doesn't lose inmediately, else if the $K$ coins are 3 coins apart from each other any of the two moves involving one of coins $K$ does not lose inmediately, if the coins are one coin apart moving them together in the avaliable direction by flipping is sufficient as there will always be two $M$ coins together so Solveig not only does notlose inmediately she can be ready anytime to flip these now two $K$ coins together back to $M$ , and finally if the two $K$ coins are 2 coins apart from each other there is always a way to make then 3 coins away from each other and this is what Solveig will do to not lose inmediately.
If at some point there is no coins $K$ then Solveig just flips 1st and 2nd coins back to $K$ and clearly does not lose in one move here.
If on the first move or later on Peer leaves Solveig a position with 3 coins $K$ then if there is two of them together at least Solveig just flips them and does not lose inmediately, else if they were all split apart this could be a starting position which is having 1st, 3rd and 5th coin flipped to $K$ , in this case just set 3rd and 5th coin together with a flip, that is what Solveig must do to note lose inmediately as the two remaining $M$ coins will be together.
If later on Peer leaves Solveig a position with 4 coins $K$ then by pigeonhole there is always two of these together so Solveig just flips them and avoids losing inmediately as the only position where there is only two $K$ coins and Solveig loses inmediately is if they were 2nd and 4th coins but there is no $M$ together here which can't happen due to how Solveig got rid of two.
Finally last thing that remains to check is that Peer does not have a move that inmediately wins, so just check his first moves: only single coin $K$ or two coins $K$ in which Solveig always has a response, the only 3 coins $K$ also has a response from Solveig as seen above, therefore for all $n \ge 5$ , Solveig can stop Peer, thus we are done :cool:

:cool:

.

This post has been edited 1 time. Last edited by MathLuis, Apr 12, 2025, 8:15 PM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a