AoPS Academy
such that 
. Find the maximum::
is called stylish if the set 
can be partitioned into 
pairs such that the sum of the elements in each pair is a power of 
. For example, 
is stylish because the set 
can be partitioned as 
, with sums , 
, and respectively. Determine the number of stylish numbers less than 
.
red balls and 
white balls. Audi took two balls at once from inside the bag. The chance of taking two balls of the same color is ...
unit. The area of the hexagon is ...
and are the roots of the cubic equation 
. The value of 
is ...
so that 
and 
is ...
, 
, 
, 
has an average of 
and a median of 
. The largest number of such data is ...
are integers greater than 
which are four consecutive quarters of an arithmetic row with 
. If 
and 
are squares of two consecutive natural numbers, then the smallest value of 
is ...
, with 
, 
and 
. The points 
and 
lies on the line segment 
. with 
and 
. The measure of the angle 
is ...
meet 
for each natural number . The value of 
is ....
provided that the difference of any two numbers at least is ...
which satisfies 
are as many as ... with 
and 
. Point lies on the side so that 
. Suppose 
is a point on the side extension 
so that 
is perpendicular to . The point lies on the ray such that 
and 
. The large angle 
is ...
consists of integers with the following properties: For every three different members of there are two of them whose sum is a member of . The largest value of is ....
with 
positive reals is ....
with 
is ....
square units. Two plots are said to be neighbors if they both have one side in common. Initially, there are a total of 
coins on the chessboard where each coin is only loaded exactly on one square and each square can contain coins or blanks. At each turn. You must select exactly one plot that holds the minimum number of coins in the number of neighbors of the plot and then you must give exactly one coin to each neighbor of the selected plot. The game ends if you are no longer able to select squares with the intended conditions. The smallest number of so that the game never ends for any initial square selection is ....
board, show that there is a 
block on the board in which at least 3 of the squares are colored red.
squares of a board be colored red. This colouring will be called a k - saturation if and only if coloring any one of the remaining squares red will result in a block of 
squares at least of which are red. Find the least such that a k - saturation exists.
red in some 
board . Suppose otherwise. Note that we cannot have more than red squares in every 
adjacent columns. Since we have columns, we can have at most 
red squares. Contradiction.
, we cannot have 
or more red squares unless it is of the following configuration:
or more red squares. In a sub-board with rows and columns, suppose we can find two adjacent red squares in a row, then consider these two columns with either the immediate right column or left column. We can have at most red squares in these columns. Hence we have at least left for columns. Grouping them into blocks of , we have a contradiction. Hence we cannot find two adjacent red squares in a row, and hence it must be
red, we have 
left for rows. Contradiction. If not, remove the top rows. We will be removing at most red. Repeat this process to get a contradiction.
sub-rectangle of the 
square. It is made up of overlapping 
block. The maximum number of red squares in the 
rectangle is because if we choose any red squares, it is easy to show that of them are part of square. Hence, since there are 
non-overlapping rectangles of this configuration, with a total of 
red squares, it leaves row of uncolored squares at the bottom. All of these are not part of any considered 
square, and hence can be colored for a maximum of 
red squares.
and 
Prove that
Where 
such that ![$f(x^2)=[f(x)]^2-2$](http://latex.artofproblemsolving.com/0/8/f/08ffecb7dcbcbe9042fde764d402bff00796e0d0.png)
such that:
Write only the answers to the questions given.
to be more exact:
in years 2002-08 time was 90' for part A and 120' for part B
red squares in 
is the best.