[/quote]
,
such that 
is a prime, 
and 
.![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
,
,
.
such that for all irrational numbers 
and 
,
be positive reals such that 
.![\[18\sum_{\text{cyc}}\frac{1}{(3-c)(4-c)}+2(ab+bc+ca)\ge 15. \]](http://latex.artofproblemsolving.com/5/d/5/5d5c47a52d2d6ea7d13ce1c2fa14a08425bbfa5d.png)
Proposed by David Stoner
,
and 
lie on sides 
and 
,
is tangent to 
at 
.
lie on side 
.
intersects the circumcircle of 
again at 
,
intersects the circumcircle of 
again at 
.
,
and 
are positive integers satisfying:
on a plane, such that the angle between the two consecutive rays is 
.
is located on the first ray. The projection of 
.
,
onto the seventh ray is 
.
.
for 
and suppose 
.
,![\[ \left( a_1^k + \frac{1}{a_1^k} \right) \left( a_2^k + \frac{1}{a_2^k} \right) \dots \left( a_n^k + \frac{1}{a_n^k} \right) \ge \left( n^k + \frac{1}{n^k} \right)^n. \]](http://latex.artofproblemsolving.com/5/a/d/5ad4ea89463088b023d252d18d9f6eb411ec9f64.png)
Y by dantx5, Gibby, jam10307, Davi-8191, Kgxtixigct, OlympusHero, icematrix2, jhu08, HWenslawski, rayfish, Adventure10, Mango247
Steve is piling 
indistinguishable stones on the squares of an 
grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions 
for some 
, such that 
and 
. A stone move consists of either removing one stone from each of 
and 
and moving them to 
and 
respectively, or removing one stone from each of and and moving them to and respectively.
Two ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves.
How many different non-equivalent ways can Steve pile the stones on the grid?
indistinguishable stones on the squares of an 
grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions 
for some 
, such that 
and 
. A stone move consists of either removing one stone from each of 
and 
and moving them to 
and 
respectively, or removing one stone from each of and and moving them to and respectively.Two ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves.
How many different non-equivalent ways can Steve pile the stones on the grid?
This post has been edited 1 time. Last edited by v_Enhance, Aug 26, 2020, 4:47 AM
.
-
-
as well. @alex31415: I don't think you should say multiset, though. Each row and column should be equal in both pilings, and order matters, so if you don't call it a function, I think you should say a tuple. I tried to fill in all the gaps in my proof, mostly by choosing extremes, but I'm not completely sure so 5-7. And I did choose 
where 
and 
with the same sums are equivalent. We use an algorithmic approach, going row by row (for the first 
rows) and transform 
to 
.
,
and 
.
,
stones in the remaining rows in that column. Then, perform the operation on the rectangle through those three points. The monovariant 
strictly decreases by 2 each time, is always even, and is equal to 0 iff 
as they have teh same row sum. Finally, after transforming row 
as well. Thus, 
equivalence classes of pilings.
.
,
by 
-
and the leftmost column be column 
.
and 
such that 
and 
.
be the number of inversions. If 
,
and 
on the squares 
be the rectangle of squares with corners 
be the set of squares outside 
stones to row 
stones to column 
.
.
stones because if it has less, then there must be other stones in row 
and 
and repeat this step with the top-left undetermined square. Eventually, only 
is undetermined. But then, 
because 
and columns labeled similarly. Let the set of 
and the 
. So we find the answer for the problem to be 
.
grid on the value of 
.
of stones and we want to get to a grid 
with the same signature. If there exists a row or column in 
in 
;
for every single column sum 
,
denote the number of stones in the 
the number of stones in the 
t
possible tuples 
.
.
and 
and let that in each column be 
.
same for the columns. so the answer is their product i.e 
such that 
.
solutions to this. Proving that any two grids with the same tuple are reachable from one another and also showing all tuples are achieved by some grid can be done through some standard greedy algorithms.
![\[\begin{array}{r} r_1+r_2+\ldots+r_n = m \\ c_1+c_2+\ldots+c_n = m \end{array} \implies \boxed{\binom{m+n-1}{m}^2}.\]](http://latex.artofproblemsolving.com/c/a/8/ca88642c2b3a3fb3b791847d5f016e59cd82f755.png)
and 
gives a valid configuration: We can fix each square starting from the top square and working from left to right then top to bottom in the top 
grid, noting that the final squares are fixed. Set each square to the lower amount of the rocks left in its row and its column.
pairs are equivalent: We again use the same order of fixing squares in the top 
.
Observe that if this sum is 
we have achieved the same config. But if we haven't achieved this configuration There must exist a square that has more squares than it should have. Then there must exist has less squares than it should have in the same row and collumn as the square that has more squares than it should have that we picked. We can operate on the rectangle containing these three squares, to give a stone to the squares which don't have as much squares as they need. This will either decrease our sum by 
,
.
ways to chose the sums of the collumns and the rows. Therefore our answer is indeed 
.
denote the stones in 
remain invariant under the operation.
.
.
are equivalent:
and at board 
.
is a permutation of 
and therefore, both of them are equivalent (as every permutation of any sequence can be reached from another other permutation by just: swapping any two elements).
such that: if there are 
stones in 
.
for all 
.
.
-
stones on the board to be 
and the invariants 
,
