[/quote]
,
is strictly increasing : 
such that for all 
. 
. 
q
answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?
,
.
Let 
.
be positive real numbers such that 
.
.
satisfying the following conditions:
divides 
,
divides 
,
.
and 
. Prove that
be positive reals such that 
.![\[\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \ge \frac{1}{3}.\]](http://latex.artofproblemsolving.com/3/b/d/3bd44cede90261ab949beabd32f8c08411ddc051.png)
such that for all 
,![\[
f(x + f(2y)) + f(x^2 - y) = f(f(x)) f(x + 1) + 2y - f(y).
\]](http://latex.artofproblemsolving.com/d/1/0/d10c01d7d416243614b24f04218d157e8b025ff2.png)
such that for all 
.
such that 
there exist 
such that 
and 
. Note : 
and 
for all 
. 
.
determine the minimum and maximum value of![\[\frac{\sum_{i,j=1}^n|a_i+a_j|}{\sum_{i=1}^n|a_i|}.\]](http://latex.artofproblemsolving.com/6/f/2/6f2427edf9a5e8e8f2e12276d0a841765e8322f1.png)
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
.
.
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 
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.
