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.
satisfies
for some 
and for all 
, then there exists a unique solution 
of the functional equation 
for all such that
for all 
.
be distinct polynomials of degree 
with non-zero coefficients. Suppose that they have 
common real roots counting multiplicity and 
common coefficients. Determine the maximum possible value of 
.
such that:
is strictly increasing,
such that 
for all 
,
, there exists 
such that![\[
f(y) = \frac{f(x) + f(x + 2024)}{2}.
\]](http://latex.artofproblemsolving.com/b/e/2/be26213154bb74bd5a35b8d160011351871bfa9b.png)
Proposed by Pavle Martinović
such that for all positive integers 
and 
,![\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]](http://latex.artofproblemsolving.com/8/b/9/8b90a9326c0fa448a5a574809c184f43513f768f.png)
Jaedon Whyte
be real numbers such that 
and 
Prove that
Where 
such that 
divides 
.
workers stand in a row. Every contiguous group of at least 
workers is a 
. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker’s number of assignments is divisible by 
. Prove that the number of such ways to assign the leaders is divisible by .
be a fixed acute-angled triangle. Consider some points 
and 
lying on the sides 
and 
, respectively, and let 
be the midpoint of 
. Let the perpendicular bisector of intersect the line 
at 
, and let the perpendicular bisector of 
intersect the lines and at 
and 
, respectively. We call the pair 
, if the quadrilateral 
is cyclic.
and 
are interesting. Prove that 
and be four prime numbers such that 
Prove that the sum of the four prime numbers is divisible by 
.
and 
and one starting number from the set 
used. Alice chooses independently her game number and Bob chooses the starting number. The other number is given to Bob. Then Alice adds her game number to the starting number, Bob adds his game number to the result, Alice adds her number of games to the result, etc. The game continues until the number 
is reached or exceeded. wins. If is exceeded, the game ends in a draw.
Show that Bob cannot win. What starting number does Bob have to choose to prevent Alice from winning?
. Let 
be the given assertion. Notice 
gives 
so either 
or 
.
gives 
. Which can be checked to work.
gives 
. The assertion gives 
. Combined we get 
. The assertion 
gives 
so either 
or 
. If then 
and 
gives a contradiction. Now assume 
. We get from 
that 
. Thus we get 
. Thus 
whenever 
is even or a square and 
can be either or 
when is not even or a square. It is easy to check that all such functions work.
, then 
from now on
then 
, contradiction, so 
, 
, so 
, and so 
, which means even numbers and perfect squares must have 
, the others can have 
as sets don't count duplicates. If it is multiset then the answer is 
.
Prove that
Prove that
Where 
satisfies
for any non-negative integers 
.
