[/quote]
,
be a positive integer. For every positive integer 
the sequence 
is defined, where 
are integers. Among these 
such that![\[
f(xy) = f(x)f(y) \;-\; f(x + y) \;+\; 1,
\quad \forall x, y \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/7/a/5/7a51a7be2306d18f4e80db1f5aac6e9bbf25c25c.png)
be a sequence of real numbers with 
.
of a permutation 
of 
to be the minima of the sum 
over all real numbers 
.
,
,
be a nonempty set of positive integers. We say that a positive integer 
is clean if it has a unique representation as a sum of an odd number of distinct elements from 
such that for any positive integer 
,
.
and 
lie respectively on sides 
and 
of a triangle 
and 
.
and 
cross at 
.
and 
cross again at point 
.
satisfying 
and their sum is an even number, consider the quadratic polynomial:![\[
P(x) = x^2 - (m^2 - m + 1)x + (m^2 - n^2 - m)(n^2 + 1).
\]](http://latex.artofproblemsolving.com/d/5/1/d51aaddfddfe06cc91abfdd3476a155770a8a5fd.png)
are positive integers but are not perfect squares.
such that![\[
ad = bc, \quad a + d = 2^k, \quad b + c = 2^m
\]](http://latex.artofproblemsolving.com/1/1/d/11d16732e3a77af46ad17adecf104a442a5d7f67.png)
for some positive integers 
and 
.
and intouch triangle 
.
onto 
.
,
respectively. Finally, let the rays 
,
meet the circumcircle of 
in 
respectively. Prove that the tangent from 
meet on the circumcircle of 
be a triangle with circum-circle 
.
be a point in the interior of triangle 
.
meet 
respectively. Suppose 
is the point of intersection of 
with 
;
is the point of intersection of 
with 
.
is parallel to 
.
is a perfect square but none of the integers 
are perfect squares.
.
and 
intersect at 
.
be the midpoint of 
.
is chosen on the extension of 
beyond 
is chosen on the segment 
such that 
.
and 
.
bisects 
.
.
be the points where the incircle touches the sides 
respectively. Let 
meet the line 
respectively. Further assume that both 
are outside the triangle 
are concyclic.
.
Y by Davi-8191, anantmudgal09, tenplusten, AlastorMoody, amar_04, SSaad, centslordm, megarnie, jhu08, mijail, Adventure10, Mango247, and 1 other user
Several positive integers are written in a row. Iteratively, Alice chooses two adjacent numbers 
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.
Proposed by Warut Suksompong, Thailand
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.Proposed by Warut Suksompong, Thailand
This post has been edited 3 times. Last edited by orl, Aug 8, 2013, 9:09 PM
Reason: Edited Title and Changed Format of The Problem.
Reason: Edited Title and Changed Format of The Problem.
to 
where 
.
has been increased by at least 
.
doesn't increase.
and hence the game has to end somewhere.
the game has ended).
(i think, it has been written by Carlos di Fiore as a comment ). Show that this cant be better.
greater than all the numbers in the original sequence. Now, read the number in base 
,
,
,
,
,
and the number would be: 
.
.
.
.
.
at all times.
is trivial. After 1 move we have 
as both are integers, so no more moves can be performed.
be negative for any 
.
,
.
.
as they are both integers, and so 
never increases.
.
,
and thus 
.
again, so 
must become 
after some (finite by the inductive hypothesis) sequence of moves. This can be reached either by 
or otherwise. Regardless, after the necessary move on the first pair, we have 
.
,
-
as 
are integers, and so 
.
.
.
and 
.
or 
.
by definition. Thus the second pair must be operated on, at which point 
.
,
,
and 
,
and 
.
.
.
be the numbers during some stage.We consider the sum 
.
.
We use strong induction on 
the base case 
numbers, there can be only finitely many operations on them. Now append another number to the list. Let the numbers written be 
and let 
Note that 
never changes. Alice cannot operate on 
and 
Hence, she must operate on 
or otherwise the desired result would follow by applying the inductive hypothesis on 
and 
We then have 
After applying finitely many operations of 
,
from left to right.
,
.
doesn't change. Consider the quantity 
.
,
it is trivial (0 steps).
denote the maximum number of steps for a sequence length 
.
number of steps to until there are no more possible steps 
such that if 
,
,
,
is to the left of 
such that if we order some set with the 
t
,
,
.
permuations, so the number of steps Alice can perform is at most 
Don't have to be adjacent.
Does not change over the iterations.
be a number written in base 
.
strictly grows over the iterations and cannot grow over 
(In base 
(from left to right) is a strictly increasing. However, we see that the maximum number can only move to its right and its moving must terminate after a certain number of moves. Thus the algorithm must terminate after finitely many iterations.
be the numbers with 
be the maximum of them.
Maximum number 
As 
Hence after each move the new two numbers are less than or equal to 
So 
increases in every step.
If 
which is true as 
If 
which is true as 
So, the number of steps are finite 
since it also increases the sum of the numbers. So, Alice must perform the second operation when 
infinitely many times – after some point, her moves will only be the second operation on 
.
in that order.
is nondecreasing.
.
.
.
.
,
to the left of 
to the number at position 
,
be the weight at 
be the weight at 
Note that both of these are greater than the initial, so our weighted sum takes an increase after each move.
is obvious.
,
.
,
,
are obvious.
of arbitrary positive integers the process terminates after 
moves and the integers are in increasing order from left to right, i.e. if the integer 
on the left end is on slot 
is on the right most slot 
.
.
of all 
written on the board.
or on 
.
can't be operated on anymore after finitely many moves (say 
and 
moves respectively) and that 
and 
.
after some time. 
,
)
or our maximum is on the right most slot.
Clearly each move takes the current permutation on some fixed direction either up or down 
(from the set of 
)
then 
,
config, notice that: 
is lexicographically smaller than 
.
from left to right. Then the number in base 
from left to right is strictly increasing. Since this number cannot go beyond 
,
