[/quote]
,
such that 
is a perfect square.
, such that for all 
:![\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]](http://latex.artofproblemsolving.com/1/4/1/1418fad9fd00b31c38bfbc5657e2a322d66ef450.png)
as the set of all numbers of the form 
for some nonnegative 
and 
.
such that 
and 
.
,
. Prove that there will be two numbers on the board after some minutes, such that one is divisible by the other.
- Greatest common divisor of numbers 
and 
board, there is initially a tree of height 
.
.
such that the gardener can ensure there are eventually 
,![\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]](http://latex.artofproblemsolving.com/7/b/6/7b6e32d32d89ca058a06974e7276f0a5874d90a7.png)
such that the equation
be a polynomial with integer coefficients such that 
.
is divisible by 
.
such that for every positive real 
,
satisfying![\[
\frac{\varphi(n!)}{n^\alpha\,(n-1)!} \;>\; c.
\]](http://latex.artofproblemsolving.com/4/4/4/4443cf7166cc65e67e2e504a826c4ac5f91aab2b.png)
be a positive integer and let 
be a strictly increasing sequence of 
be a subset of 
such that the value of![\[
\left|1-\sum_{i \in X} a_{i}\right|
\]](http://latex.artofproblemsolving.com/2/5/1/2515e074de1782f38453e1fce8ab7c2ec7cfe4c0.png)
is minimised. Prove that there exists a strictly increasing sequence of 
with sum equal to 2 such that![\[
\sum_{i \in X} b_{i}=1.
\]](http://latex.artofproblemsolving.com/3/7/4/37410820fc4c546f11d57212cfce53a7ec4941e3.png)
be a scalene triangle with circumcircle 
,
at 
.
meets 
and 
.
intersects the 
-
,
,
.
passes through either 
be positive real numbers such that : 
. Prove that : 
![$P, Q \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/c/3/0/c301578bade5b93056d77454921cba0a9812212c.png)
be relatively prime nonconstant polynomials. Show that there can be at most three real numbers 
such that 
is the square of a polynomial.
Y by Amir Hossein, Davi-8191, rustam, Kayak, Tenee, Purple_Planet, centslordm, jhu08, megarnie, Adventure10, Mango247, lian_the_noob12, Sedro
A rectangle 
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of are either all odd or all even.
Proposed by Jeck Lim, Singapore
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of are either all odd or all even.Proposed by Jeck Lim, Singapore
This post has been edited 2 times. Last edited by MarkBcc168, Jul 15, 2018, 12:57 PM
.
board. Color 
and 
with 
,
if 
,
if 
.
.
.
,
.
is even and so 
is odd. Repeat in the other direction (up and down instead of left and right).
,
tile it with dominoes and for 
mark the leftmost corner and tile the remaining rectangle with dominoes. Observe that each tiling 
maps to a tiling 
consisting only of dominoes and boxes such that 
white if 
,
squares have at most 2 corners.
,
be the side of one other small rectangle as well?
are both even or odd; we also know for sure that the distance from the horizontal sides of this rectangle to 
![$[0,a]\times [0,b]$](http://latex.artofproblemsolving.com/6/0/1/601076a1c0a357d6bd2c71085669c8c3074b0129.png)
and we are considering ![$[x,x+c]\times [y,y+d]$](http://latex.artofproblemsolving.com/3/6/b/36b9f1541a35e6b750fd01d3ff4dbb3765c805c2.png)
,
,
are odd and 
is possible. Let us just focus on the first half. We can note that by weighting each cell by 
,
must be odd, so there must be some rectangle with 
odd.
.
.
,
black corners. Iff the rectangle's distances from the four sides are all odd or all even, then it has 
white and 
,
in the dissection has the following property: the distance from 
,
to be the number of black squares contained in it, and 
to be the number of white squares in it. Then the condition on our required rectangle 
.
for every rectangle. Then 
where 
refers to the rectangle covering the entire of 
refers to the partition. But this is impossible due to 
,
Such a small rectangle would have odd area and thus odd side lengths, meaning the distances from each pair of parallel sides have the same parity (i.e. the distance from the top and bottom would both be odd or both be even)
,
,
and 
rectangles. Suppose such a rectangle doesn't exists. Then, for any rectangle 
which is contained in some division of 
If a smaller rectangle 
and 
or in a 
and 
will never generate rectangles with that.
By a 
rotation, it suffices to consider the first case. Note that the piece 
squares. Color the cells chessboard-style with the corners all white and number the white cells 
with even area
with even area
with even area
with odd area
with odd area
with odd area
![\[
f(\mathcal{D}) = \int\int_\mathcal{D} e^{\pi i(x+y)}
\]](http://latex.artofproblemsolving.com/7/2/e/72e842bd369b5836cf555d4b33a018fc33ed45fd.png)
is nonzero iff it has all odd side lengths. Since 
is nonzero, the 
of one of the rectangles in the partition must also be nonzero.
