[/quote]
,
and 
.
.
of degree 
.![$[a;b]$](http://latex.artofproblemsolving.com/0/0/6/0066f4b2714f88cdaf883b00f877aa1355b7f9ef.png)
and Maxim will give in response the maximum value of 
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v_p(n!) \leq \frac{n}{p - 1}
\]](http://latex.artofproblemsolving.com/5/7/c/57ca4d0a6532fb4b97687a97b9c1fb35ad02b9f2.png)
and 
Prove that
Let 
and 
Prove that
not isosceles, the incircle 
.
is the intersection of 
.
is the projection of 
onto 
.
cut 
.
is the projection of 
onto 
.
cut 
respectively. Prove 
are collinear
be the set of non-negative integers and 
be the set of positive real numbers. Let 
be a function such that 
and 
for all integers 
,
for all integers 
.
.
which satisfy 
![$ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$](http://latex.artofproblemsolving.com/c/d/a/cdae05ad34239366b0dd3921243522c711b6ae58.png)
method, I can assum 
![$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $](http://latex.artofproblemsolving.com/c/d/4/cd481672bee1e87d8c97ac623436e0ec75cfa73a.png)
are given such that 
.
in natural numbers m and n.
denote the smallest possible value of 
where 
are sets such that 
and 
whenever 
.
and 
intersect at 
at 
respectively. Let 
be the symmetric points of 
intersects circles 
respectively which are distinct from 
and 
are equal.
,
.
intersects 
at 
.
cuts segment 
at 
respectively. If 
and 
,
equals 
.
are points on a circle 
,
,
.
at 
is on 
such that 
.
intersects 
.
is the reflection of 
.
lies on the same circle.
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 
