[/quote]
,
such that 
is divisible by 
,
is divisible by 
,
is divisible by 
.
?
and a positive integer 
is given.
and for each 
![\[f_n(k)=\frac{a_n+\dots+a_{k+1}}{n-k}-\frac{a_k+\dots+a_1}{k}\]](http://latex.artofproblemsolving.com/0/4/1/041f80cf63f17c0f62c166021af474a5d305eb87.png)
Then for every integer 
the condition![\[a_{n+1}=\max_{0\leq k<n}(f_n(k))\]](http://latex.artofproblemsolving.com/b/1/f/b1f5aff787de8fc9c113cf8441a6b2900a19f24f.png)
is satisfied. Prove that this sequence must be eventually constant.
be a positive integer. We are given a 
board whose unit squares are colored in black and white in such way that starting with the top left square, every third diagonal is colored in black and the rest of the board is in white. In one move, one can take a 
square and change the color of all its squares in such way that white squares become orange, orange ones become black and black ones become white. Find all 
![$Q\in \{-1,1,0\}[x]$](http://latex.artofproblemsolving.com/b/4/0/b404c8d6415aaf3d4442cc5381f582c1f8f8b996.png)
with 
.![$\exists P \in \mathbb Z[x]$](http://latex.artofproblemsolving.com/a/2/e/a2e67dd74df9971ca90d1a5d81c1c8e1f3982276.png)
palindrom with 
?
to 
and city 
has a bidirectional road of lengths 
if and only if 
and 
is a prime number.
has incentre 
and incircle touching 
at 
.
be the antipode of 
in the circumcircle of 
is chosen on the internal angle bisector of 
such that 
.
,
be the midpoint of 
.
for which:
for all 
.
particles on a circle situated at the vertices of a regular 
be the smallest number of collisions after which all particles return to their original positions. Find 
for which 
,
denote the integer part.
and 
Prove that
chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board.
of 
integers satisfying the following condition: for each 
,
such that 
is divisible by 
.
Y by laegolas, Tumon2001, Adventure10, Mango247
General Tilly and the Duke of Wallenstein play "Divide and rule!" (Divide et impera!).
To this end, they arrange
tin soldiers in 
companies and command them by turns.
Both of them must give a command and execute it in their turn.
Only two commands are possible: The command "Divide!" chooses one company and divides it into two companies, where the commander is free to choose their size, the only condition being that both companies must contain at least one tin soldier.
On the other hand, the command "Rule!" removes exactly one tin soldier from each company.
The game is lost if in your turn you can't give a command without losing a company. Wallenstein starts to command.
a) Can he force Tilly to lose if they start with
companies of tin soldiers each?
b) Who loses if they start with
companies consisting of 
tin soldiers?
To this end, they arrange
tin soldiers in 
companies and command them by turns.Both of them must give a command and execute it in their turn.
Only two commands are possible: The command "Divide!" chooses one company and divides it into two companies, where the commander is free to choose their size, the only condition being that both companies must contain at least one tin soldier.
On the other hand, the command "Rule!" removes exactly one tin soldier from each company.
The game is lost if in your turn you can't give a command without losing a company. Wallenstein starts to command.
a) Can he force Tilly to lose if they start with
companies of tin soldiers each?b) Who loses if they start with
companies consisting of 
tin soldiers?
moves, so the last move would be performed the Duke, which would imply that the Duke has won.
,
)
is odd, and there are 
companies consisting of 
