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 real positive numbers. Show that ![\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1\]](http://latex.artofproblemsolving.com/0/9/5/0959464bbb9b821a11df292f9e6addded7326371.png)
+1 w
is almost squarefree if there exists a prime number 
such that 
and 
is squarefree. Prove that for any almost squarefree positive integer 
the ratio 
is an integer.
be a natural number. At a mathematical olympiad training camp the same 
courses are organised every day. Each student takes exactly one of the courses each day. At the end of the camp, every student has takes each course exactly once, and any two students took the same course on at least one day, but took different courses on at least one other day. What is, in terms of , the largest possible number of students at the camp?
such that 
for all 
?
have the same incenter 
.Prove that 
is concyclic iff 
is concurrent
and be positive integers. Suppose that the sum of integers between and , including and , are equal to 
.
, find the pair such that achieves the minimum.
be real numbers such that 
and
be integers such that, for each 
, there exists a subset 
with![\[
\sum_{j \in S} a_j \equiv i \pmod{2048}.
\]](http://latex.artofproblemsolving.com/8/b/e/8be39aaf83795f611b1c3fa0d5971905bd6ebdfe.png)
Show that for each 
, there is exactly one 
such that 
is divisible by 
but not by 
.
is the summation notation, for instance, 
, while for the empty set 
, one defines 
.
, 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 
.
.![$\sqrt[3]{ab}+ \sqrt[3]{cd} < \sqrt[3]{(a+b+c)(b+c+d)}$](http://latex.artofproblemsolving.com/0/5/0/0502ba65f0545b3af84c41b2549d6256550b473b.png)
.
, such that for all 
:![\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]](http://latex.artofproblemsolving.com/1/4/1/1418fad9fd00b31c38bfbc5657e2a322d66ef450.png)
such that for any positive integer 
there exists a positive integer 
are equal to 