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.
, find the minimal integer 
, such that for any positive integer 
, in any subset with elements of the set 
there are at least 
relatively prime elements.
is polynomial with minimum degree such that 
for 
, find 
and 
such that 
and is to the left of , and replaces the pair 
by either 
or 
. Prove that she can perform only finitely many such iterations.
such that 
and 
are both perfect squares and 
is a power of a prime number.
distinct points in a plane such the circumradius of any of them is the same.
Find the number of orderings of the students such that the shortest person
Define the badness of an ordering as the maximum number 
such that there many people with height greater than in front of a person. For example:
has badness since there are numbers greater
in front of it. Let 
denote the number of orderings of 
with badness . Find .
is a sequence of real numbers with 
such that for all 
Prove that
is call 
if it satisfies the following properties:
and 
for all indices 
.
. for which: Every sequence, there always exist two terms whose diffence is not less than . (where is given positive integer)
such that 
.
Prove that :
Where 
Prove that
Where 
as the number of orderings of n with badness less than
Prove that 
Where 
be a positive integer. Consider the following equation: ![\[ \{x\}+\{2x\}+ \dots + \{nx\} = \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2nx \rfloor\]](http://latex.artofproblemsolving.com/8/2/9/8299b3ceb1650418c0a373536cb1d2931900efd9.png)
a) For 
,
.
real solutions.
