[/quote]
,
such that 
.
.
which satisfy the condition![\[ f(m + n) \geq f(m) + f(f(n)) - 1
\]](http://latex.artofproblemsolving.com/4/b/d/4bddce124cf0da82cb8bb6c6ae2325239354f27c.png)
Find all possible values of 
.
be a positive integer and 
be a positive integer coprime to 
.
,
,
Find, in terms of 
,
for which there exists an index 
such that 
is divisible by 
.
table are initially white. Alice and Bob play a game. First Alice paints 
rows and 
be an interior point of the acute triangle 
with 
so that 
The point 
on the segment 
satisfies 
the point 
on the segment 
satisfies 
and the point 
on the line 
Let 
and 
be the circumcenters of the triangles 
and 
respectively. Prove that the lines 
and 
are concurrent.
,
as the product of the digits of 
.
be a function satisfying the following conditions:
,
and 
,
;
is surjective.
.
and 
.
and 
respectively. The point 
such that 
is equilateral. Let 
and 
.
.
.
,
.
denote the number of ways this can be done. For example, clearly 
because we can have either a red or a blue tile. Also 
since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.
for all 
.
for all 
.![\[ \binom{m}{r} = \begin{cases}
\dfrac{m!}{r!(m-r)!}, &\text{ if $0 \le r \le m$,} \\
0, &\text{ otherwise}
\end{cases}\]](http://latex.artofproblemsolving.com/b/8/d/b8d653661e421db1fce7d93837e7bce31337bb30.png)
for integers 
.
where 
for some non-negative 
such that
for every 
.
be real numbers such that ![\[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]](http://latex.artofproblemsolving.com/3/d/0/3d015292c249a69a44d5f923092201d7a271b896.png)
Find the smallest possible value of 
.
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 
.
,
is unchanged. But (*) cannot be performed infinitely many times.
Don't have to be adjacent.
be the initial numbers on the row.
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 
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 
is finite and thus Alice can only perform a finite amount of moves. QED
(from the set of 
)
then 
but 
,
config, notice that: 
is lexicographically smaller than 
.
permutations, it should take at-most 
from left to right. Then the number in base 
from left to right is strictly increasing. Since this number cannot go beyond 
,
of 
such that the number in the 
-
,
.
if, at the first value where 
.
,
,
and 
,
,
.
from our total board state, this "shifts" all values corresponding to numbers greater than or equal to 
.
steps, as desired.
bound is weird, so it probably relates to 
.
on its right side, in which case we cannot make any move.[If 
and after making a move we can get 
,
,
.
with either of the moves the larger number is always to the right hence no other replacement can be made to this pair.
and 
and 
which will both be sorted in a finite amount of moves hence no other replacement besides the one moving M to the right. Now this argument can be repeated until M is at the rightmost place in the sequence. Now we can ignore M from the sequence since no moves can be done with it and by the inductive hypothesis the rest can be sorted as well.
