AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
ones are written on a blackboard. Grisha and Gleb are playing game, by taking turns; Grisha goes first. On one move Grisha squares some 
numbers on the board. On his move, Gleb picks a few (perhaps none) numbers on the board and increases each of them by 
. If in 
moves on the board a number divisible by 
appears, Gleb wins, otherwise Grisha wins. Which of the players has a winning strategy?
be a nonempty subset of the points in the Cartesian plane such that for each 
exactly one of 
or 
also belongs to . Prove that for each positive integer 
there is a line in the plane (possibly different lines for different ) which contains at least points of .
such that for all positive real numbers 
,![\[
f^2(x) \geq f(x+y)\left(f(x) + y\right).
\]](http://latex.artofproblemsolving.com/7/1/f/71fb0d51492ee4c3b2bf4049e8224882fec16020.png)
such that 
is a divisor of 
.
players of a hockey team gather to select their team captain. Initially, they stand in a circle, and each person votes for the person on their left.
updates their vote, one at a time, according to the following procedure: At the time of the update, if is voting for 
and 
is voting for 
then updates their vote to 
(Note that 
and 
need not be distinct; if 
then 's vote does not change for this update.) Every player updates their vote exactly once in each round, in an order determined by the players (possibly different across different rounds). rounds. Prove that at this time, all players will unanimously vote for the same person.
be a positive integer that is not divisible by 
. Furthermore, let 
be a positive integer. Prove that none of the numbers of the form 
is a square number.
, 
, 
, 
lie on a semicircle 
in this order with diameter 
, and is not parallel to 
. Points 
and 
lie on segments 
and 
respectively such that 
and 
. A circle 
passes through and is tangent to , and intersects again at 
. Prove that the lines 
, and 
are concurrent.
if the following conditions are satisfied: 
, 
, 
.
be a triangle with incircle touching 
at 
respectively. Let 
and 
be its circumcenter and midpoint of 
Suppose that circumcircles of 
and intersect at for the second time. Assume 
is on the circumcircle of such that 
is cyclic. Prove that circumcenter of 
lies on 
satisfy 
. The circumcircle 
and the incircle 
of are tangent to the sides 
at 
, respectively. The line 
meets 
at and intersects at 
, while the line 
meets 
at 
. Construct the common external tangent 
(different from ) to the incircles of the triangles 
and 
. Show that is parallel to the line 
.
are being replaced with 
. Find the biggest number of steps, after which the number 
will appear on the board.
.
, because last four digits would have to be divisible by 
, but in each of this cases numbers 
are not. Similarly, we can see that 
leads to contradiction. is coprime to . We know that 
is coprime with 
and 
for all , so function 
is bijective on integers relatively prime to 
(we count modulo ). To see that this function is bijective, we can also see that 
iff 
divides 
, but cannot divide neither 
, nor 
if 
is not divisible by , so we would have to have 
. The same argument works modulo 
. Therefore, we can have 
. Obviously, we can also have 
, so our answer is 
.
,
,![\[ f(x)f(yf(x))=f(x+y). \]](http://latex.artofproblemsolving.com/5/6/0/560049dd70944cafa0423e5f1229d215f78215a0.png)
such that for any positive integer 
there exists a positive integer 
are equal to 
