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.
marbles in 
colors such that there are exactly 
marbles of each color. She wants to arrange them on 
shelves, 
marbles on each shelf such that for any 
colors there is a shelf that has marbles of those colors.
be an integer. The numbers 
are split into 
groups of . Prove that there are at least 
quadruples 
such that they are all in different groups, 
and 
.
be positive real numbers so that 
. Prove that![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
be an acute triangle with orthocenter 
. Prove that there is a line 
which is parallel to 
and tangent to the incircles of 
and 
.
are three positive real numbers, prove that 
be positive integers. Prove that it is impossible to have all of the three numbers 
to be perfect squares.
be a positive integer. A positive integer is called a -good number if it satisfies has exactly 
digits in decimal representation (it cannot have leading zeros). digits and the last digits of are considered as two separate -digit.
is a -good number because![\[(20 + 25)^2 = 2025.\]](http://latex.artofproblemsolving.com/5/c/5/5c51bca485cea6c3750c48c09091eb5fd72b5192.png)
Find all 
-good numbers.
be the set of all positive integers greater than or equal to 
and let 
be a function such that 
for all 
if 
, determine 
players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least 
next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
be a fixed positive real number. Find the minimum possible value 
such that for any positive reals 
, 
, 
, 
, we have
and with no common divisor greater than 1 such that 
divides 
. (Professor Yongjin Song)
and 
. Prove that there always exists a positive integer 
such that 
. is a power of a prime number, but for arbitrary then well.....
, 
, 
, 
are integers such that 
.
and 
of 
such that for each integer 
with 
, we have![\[ n\mid a_i - b_i - c_i
\]](http://latex.artofproblemsolving.com/1/a/e/1ae8d54f068b646ac7a1e43d75e7f26510cfdc95.png)
be a finite abelian group of order .
be arbitrary.
and pairwise distinct 
such that 
for 
.
and 
s.t 
and 
and there are exactly two i;j s.t 
and 
.Then if we know two permutation good for the sequence (a_1...a_n) then we can build two permuttionm good for (b_1...b_n)
Please post it here and now!
that add 
to 
and subtract from 
.
that add to and subtract from .
and 
of ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
for some sequence ![$\{ a_i\}_{i\in [n]}$](http://latex.artofproblemsolving.com/f/6/e/f6e384c04b4aaf07513dacb66f8cadc0a597069e.png)
so that 
for all ![$i\in [n]$](http://latex.artofproblemsolving.com/e/e/5/ee51c5c5eb09111cc774a00e9970074ad9f9058c.png)
.![$\{ b_i\}_{i\in [n]}$](http://latex.artofproblemsolving.com/1/3/e/13e91e327fa3f2bc2d43edbe110540d350184040.png)
with sum divisible by that differ, in modulo , from in only two positions, say 
and 
.
and 
of so that 
for all .
for all that 
.
by, for each integer 
, define ![$i_{k+1}\in [n]$](http://latex.artofproblemsolving.com/c/e/5/ce54cff3726a2dae0e8cc97f7b358dd90622aeee.png)
to be the unique integer satisfy 
.
are the indices that 
with minimal 
, and then minimal 
.
. This means 
for all 
. for 
gives us
Plugging and use 
gives us 
.
, contradiction to the definition of 
.
. Let 
. Define the desired permutations and as follows:
and 
for the rest that 
.
is just to not use 
more than one time.
for all except when 
. and are permutations of , we've ![$\sum_{i\in [n]}{(\sigma' (i)+\tau' (i))} \equiv_n 2\times \frac{n(n-1)}{2}\equiv_n 0\equiv_n \sum_{i\in [n]}{b_i}$](http://latex.artofproblemsolving.com/4/0/c/40ce5067c15701668f9fc9a82369ce8ab1f4384c.png)
.
when too. This prove the validity of permutations we constructed.
for 
on the coordinate plane with coordinates modulo . We want to show that it is possible to choose the points plotted to have 
be any multiset of elements 
such that sum of the elements is 
. Note that the points plotted can be any points not in the same row or column.
points not in the same row or column such that the multiset 
of 
for all of the points 
can be any multiset of elements . Then, we can choose the unique point not in the same row or column as the points to obtain points for the original statement, as the sum of the coordinates of all points is necessarily 
. points in the following way. arbitrary points not in the same row or column. We will change the points in such a way that we can replace any element of by any residue . Suppose we change the element in corresponding to a point 
. Then, we will shift horizontally until the sum of the coordinates of is the desired value . Now, we might have two points in the same column. We will repeatedly perform the following operation. distinct rows taken up by the points. If the last point moved was and it is in the same column as 
, we will move on the line through with slope 
until occupies the previously empty row. to a point 
, then on the next step it moves to a point 
. We see that 
and 
both have slope . Now, note that and must be in the same row. Furthermore, after the operation moves to , the row containing must be empty, meaning that must be moved to that row. This implies that and are in the same row. Therefore, we see that 
is a cyclic isosceles trapezoid and in particular the perpendicular bisectors of and are the same, proving the claim. points to distinguish them. Now, there are finitely many states of the points, so eventually the operation must loop through a cycle of states. In this cycle, start at an arbitrary state. Now, since at least 
point must be moved and then return back to its original position, consider the point which returns to its original position the earliest. When was moved, the next point which was moved must have occupied the original row of . However, if was moved back to its original position, then the original row of must be vacant, so must have been moved. However, this implies that was moved back to its original position before by the claim, a contradiction, so we are done.
.
