\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{Kosovo TST 2019/1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 106}
\maketitle
\section*{Problem}
There are $2019$ cards in a box.
Each card has a number written on one of its sides
and a letter on the other side.
Amy and Ben play the following game:
in the beginning Amy takes all the cards,
places them on a line and then she flips as many cards as she wishes.
Each time Ben touches a card he has to flip it and its neighboring cards.
Ben is allowed to have as many as $2019$ touches.
Ben wins if all the cards are on the numbers' side, otherwise Amy wins.
Determine who has a winning strategy.
\section*{Video}
\href{https://www.youtube.com/watch?v=1HWZAbOob44&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/1HWZAbOob44}}
\section*{External Link}
\url{https://aops.com/community/p12173180}
\newpage
\section*{Solution}
Ben wins.
First, note: $2019$ doesn't matter because moves commute with each other
and repeating a move does nothing.
So we're going to ignore the requirement of at most 2019 touches.
We now give an algorithm to flip any individual single card,
which is obviously sufficient.
Number the cards $1$, \dots, $2019$ from left to right.
\begin{itemize}
\ii Touch cards $1$, $2$, $4$, $5$, $7$, $8$, \dots, $2017$, $2018$
will flip only the card $2019$.
\ii Touching $2019$ in addition to the previous algorithm
will flip only the card $2018$.
\ii If we do the algorithm to flip just $n$ and $n-1$,
and then touch $n-1$, that's the same as flipping just $n-2$.
\ii Repeating this way, we get an algorithm to flip any individual card.
\end{itemize}
\begin{remark*}
In linear algebra terms, the problem can also be phrased as saying that
\[
\det
\begin{bmatrix}
1 & 1 & 0 & 0 & 0 & \dots & 0 & 0 & 0 & 0 \\
1 & 1 & 1 & 0 & 0 & \dots & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 & 0 & \dots & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 1 & \dots & 0 & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots
& \ddots & \vdots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & 0 & \dots & 1 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & \dots & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & \dots & 0 & 0 & 1 & 1 \\
\end{bmatrix}
\neq
0
\]
in $\FF_2$.
\end{remark*}
\end{document}