Ben has a winning strategy. The problem is equivalent to: Is it possible to transform an arbitrary binary array of size 2019 into a binary array of size 2019 with all ones with at most 2019 moves? We will prove that it is possible for all N, where N = 3K + 1 or N = 3K by induction. For n = 3 it is trivial. Suppose it is possible for N=3K. We will prove it is possible for N=3K + 1 Transform the first 3K elements into ones. If the last one is 1, the proof is complete. Else, the last one is zero, so if we flip the last card we will have: 111...11101 Where the first 3K-2 elements is 1. If we flip the first card we will have: 00111....1101 and the number of ones "in the middle" is 3(K-3) so we can flip all the cards there into zeroes and we will have: 000...001 and because the number of zeroes is divisible by 3 we can flip them into ones which completes the proof. We proved that we can get to an array of all ones in some number of moves ( that number may be greater than 3K+1) but because nothing changes if we flip the same card twice, it will always be possible in 3K+1 flips. We will now prove if it is possible for 3K, and 3K+1, it is possible for 3(K+1): First, flip the first 3K+1 into ones (we can do that because of the claim) and if the last two elements are 00 or 11 the proof is complete because we can flip 00 into 11. Else, the last two elements are 01 or 10. If it is 01 flip it into 10 and after that flip the first card. We now have: 00111.....10 Where the number of ones is 3K, so we can flip all the ones into zeroes and have an array of zeroes which can be transformed into an array of ones by flipping every 3X+2 card. Similarly for the 3K+1 case, it will always be possible in 3(K+1) moves. With this the proof is complete.

This problem was proposed by me. An interesting question would be if instead of $2019$ is any positive integer $n$ in that case if $n-2$ is not divisible by $3$ then Ben wins, otherwise Amy wins.

dangerousliri wrote: This problem was proposed by me. An interesting question would be if instead of $2019$ is any positive integer $n$ in that case if $n-2$ is not divisible by $3$ then Ben wins, otherwise Amy wins. Are you sure about that I think $Ben$ always wins, except for the $2$ case.

Hamel wrote: dangerousliri wrote: This problem was proposed by me. An interesting question would be if instead of $2019$ is any positive integer $n$ in that case if $n-2$ is not divisible by $3$ then Ben wins, otherwise Amy wins. Are you sure about that I think $Ben$ always wins, except for the $2$ case. Yes. Just try this example when $n-2$ is divisible by $3$. First card is on side of the letter and others are on numbers side.

let each configuration of sides of numbers and sides of letters be represented as a string of $1$s and $0$s let the set of all this string be $A$ lets define the group $A^{\times }$ as the group on set $A$ with the operation $\times$ defined as for $a_1a_2...a_{2019}$ and $b_1b_2...b_{2019}$ we have that $b_1b_2...b_{2019}\times b_1b_2...b_{2019}=c_1c_2...c_{2019}$ were $\forall i ,0<i<2020$ we have if $b_i=0 \Rightarrow c_i=a_1$ if $b_i=1\Rightarrow c_i\neq a_i $ and $c_i\in\{0,1\}$ this group clearly works as it has a identity $0000...0000=e$ it has an inverse for $a \in A ,a\times a =e$ it is clear to see its abelian so its also comutitive. note that the set $T=\{10...0,010..00,...,00....01\}$ is a generator of $A^\times$. it is clear that ben flipping a card and a neighbor is equivalent to multiplying the previous configuration by $00...01110...000$ let $B=\{110...00,1110...000,01110...00,...,00...01110...00,...,00...0111,00...011\}$ we will show that this set is also a generator, $110....00\times1110...00=001.....000000$ $a_1=1110....00\times01110...00=10010.....000000=a_2$ $a_2\times0001110...00=1000110000...=a_3$ $a_3\times00001110...00=1000110000...=a_4$ we define $a_n$ recursivly in the same way we defined $a_1,a_2,a_3,a_4$ (i lack the LaTeX skills to wright that properly) eventually, there will be an $a_i$ such that $a_i=10...011$ this is true since$ 3\mid2019 (my lack of LaTeX skills also limits this explanation)$ but we have that $a_i\times00...011=10...00=c_1, let c_3=0010...00$ $1110...00\times c_1 \times c_3= 010...00=c_2$ we define c_k similarly to $c_1,c_2,c_3$ (note that each element of $B$ is of the form $c_ic_{i+1}c_{i+2}$ except $c_1c_2$ and $c_{2018}c_{2019}$ assume that $c_1,c_2,c_3,...c_k$ can be achieved $c_{k-1}\times c_k \times c_{k-1}c_kc_{k+1}=c_{k+1}$ and also the fact that $|B|=2019$ implies $B$ is e generator but note that since each element of $A^{\times }$ has order 2 this implies every element $A^{\times }$ can be achieved by multiplying at most $|B|=2019$ combinations thus this implies Ben can reach his configuration in at most 2019 moves thus ben wins.

Solution from Twitch Solves ISL: 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. Touch cards $1$, $2$, $4$, $5$, $7$, $8$, \dots, $2017$, $2018$ will flip only the card $2019$. Touching $2019$ in addition to the previous algorithm will flip only the card $2018$. If we do the algorithm to flip just $n$ and $n-1$, and then touch $n-1$, that's the same as flipping just $n-2$. Repeating this way, we get an algorithm to flip any indvidual card. 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 ${\mathbb F}_2$.

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