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.
be a positive integer, and let 
be a subset of 
. An -partition of into 
parts is a representation of n as a sum 
, where the parts 
belong to and are not necessarily distinct. The number of different parts in such a partition is the number of (distinct) elements in the set 
.-partition of into parts is optimal if there is no -partition of into 
parts with 
. Prove that any optimal -partition of contains at most ![$\sqrt[3]{6n}$](http://latex.artofproblemsolving.com/2/0/1/201489b1f95ad200036698a606f712da7bb465b9.png)
different parts.
other knights where friendship is mutual. Prove that King Arthur can place some of his knights around a round table in such a way that every knight is friends with the 
people adjacent to him and between them there are at least 
friendships of knights that are not adjacent to each other.
cities in a country, where 
is an integer. Some pairs of cities are connected by direct (two-way) flights. For two cities and 
we define:
A 
between and as a sequence of distinct cities 
, 
, such that there are direct flights between 
and 
for every 
;
A 
between and as a path between and such that no other path between and has more cities;
A 
between and as a path between and such that no other path between and has fewer cities. and in the country, there exist a long path and a short path between them that have no cities in common (except and ). Let 
be the total number of pairs of cities in the country that are connected by direct flights. In terms of , find all possible values 
, 
.
.
are fixed positive integers.
are fixed positive real numbers.
be the set of all functions 
with the following properties: is defined for all real numbers and takes only real values.
the following equality holds: 
such that
,
.
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](http://latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.
be a cyclic quadrilateral in which the lines 
and 
meet at a point . Let 
be the point of the line 
, different from , such that 
. We construct the parallelograms 
and 
. Prove that the points 
lie on the same circle.
defined on the set of all positive integers and taking non-negative integer values, satisfying the three conditions: 
for at least one ; 
for every positive integers 
and 
; there are infinitely many positive integers such that 
for all 
.
be positive real numbers such that 
. Find the minimum value of the following sum :
knowing that the denominators are positive real numbers.
point 
was chosen such that 
and 
. Prove that if perpendicular from to passes from the intersection of diagonals of , then 
.
balls to 
boxes as she wishes. After that, Aslı and Zehra make alternated moves which consists of taking a ball in any wanted box starting with Aslı. One who takes the last ball from any box takes that box to herself. What is the maximum number of boxes can Aslı guarantee to take herself regardless of Zehra's moves?
, 
. The internal angle bisector of 
and the external angle bisector of intersect the ray at points 
and , respectively. Given that 
, prove that 
.
![\[f(xf(y) + x) = xy + f(x)\]](http://latex.artofproblemsolving.com/9/9/f/99f580ebc50846e6bc2c004667559922749a4dfa.png)
be a sequence of perfect squares such that 
can be obtained by concatenating a digit to the right of 
. Determine all such sequences that are of maximum length.
be a sequence of perfect squares such that can be obtained by concatenating a digit to the right of . Determine all such sequences that are of maximum length.
can be 
, 
can be only 
, 
can be only 
, 
can be 
only. We can't extend this upto 
as there are no such perfect squares. Done!..... though I don't know if 
can be included as a perfect square in the sequence. If can't be included then just stop it before .
.
such that:
