has been partitioned into disjoint pairs 
)
,
equals 
or 
.![\[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \]](http://latex.artofproblemsolving.com/0/4/4/04422a7d2266b92a6b3f215b698f078cda85091f.png)
ends in the digit 
.
such that 
for all positive integers 
.
be a triangle with 
Line bisector of 
intersects 
at point 
.
.
.
Let 
.
Let 
.
Let 
with edges 
.
three perpendiculars to the three lateral faces.
and radius 
.
.
exist that satisfy 
for all real 
and the domain of f is strictly within the set of 
,
?
with edge 
and tangent to the plane 
.
)
Y by Rounak_iitr
Alice and Bob are playing game on an 
grid. Alice goes first, and they take turns drawing a black point from the coordinate set
![\[\{(i, j) \mid i, j \in \mathbb{N}, 1 \leq i, j \leq n\}\]](//latex.artofproblemsolving.com/4/3/f/43f7493d1dff43079eb5481ee17bc8f16d7b9553.png)
There is a constraint that the distance between any two black points cannot be an integer. The player who cannot draw a black point loses. Find all integers 
such that Alice has a winning strategy.
Proposed by chengbilly
grid. Alice goes first, and they take turns drawing a black point from the coordinate set![\[\{(i, j) \mid i, j \in \mathbb{N}, 1 \leq i, j \leq n\}\]](http://latex.artofproblemsolving.com/4/3/f/43f7493d1dff43079eb5481ee17bc8f16d7b9553.png)
There is a constraint that the distance between any two black points cannot be an integer. The player who cannot draw a black point loses. Find all integers 
such that Alice has a winning strategy.Proposed by chengbilly
This post has been edited 1 time. Last edited by chengbilly, Mar 5, 2025, 5:09 AM
.
.
then
is not a square. Alice can mark 
since points are symmetrical with respect to 
i
.
then Bob should mark 
.
is not a square.
.
then Bob wins.
Alice can play any point, but follow the rule; then she'll win. And if 
Bob can play any point, but follow the rule; then he'll win. Cuz the problem is equivalent to: pick any point and draw all point on its row and column by black. The game must end after 
both can't be marked.
then notice 
is a perfect square
of the grid denote cell 
.
,
must be either already colored black or an integer distance away from a black point. The former is impossible due to the symmetric nature of the moves until this point. For the latter to be possible the violating pair must be either 
, 
or 
where 
.
implying that Bob performed an illegal move in his previous move. Further if 
.
must be a violating pair where 
is the reflection of 
across the center square. But by the nature of the algorithm, 
, 
,
and 
.
for 
must be a violating pair. In the later case, the reflection of 
for odd positive integers 
and 
which implies that the quantity under the square root is 
which can never be a perfect square. Thus, this distance is not an integer and hence it cannot be a violating pair. Thus, as long as Alice has a legal move, so does Bob and since the number of cells on the board is even, it is Alice who runs out of legal moves first.
