AoPS Academy
and 
lie consecutively on a line parallel to the edge of the board, then the Cannon can jump over to capture , landing on 's original position.
such that for any integers 
, no matter where the Cannon is initially placed on an 
chessboard (with all other positions occupied by pieces), it is guaranteed that after a finite number of moves, at most pieces will remain on the board.
of positive integers such that ![\[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]](http://latex.artofproblemsolving.com/3/5/3/353bbb15bac205053c26208af8cac7c32a296f55.png)
Proposed by Gabriel Chicas Reyes, El Salvador
, let 
be the incenter. Denote 
as the intersection of 
and the perpendicular bisector of 
, 
as the intersection of 
and the perpendicular bisector of 
, and 
as the intersection of 
and the perpendicular bisector of 
.
, 
, and 
are coaxial.
such that 
divides the product 
.
rectangle. At least one of the titles is an S tile. What are the other two tiles?
is just ![$[\sum_{i=1}^{n}\sigma_0(i)]-n=\sum_{i=1}^{n}[\sigma_0(i)-1]$](http://latex.artofproblemsolving.com/4/5/5/4555f8d23420fe2d9db1d8b5c288ef33ade9f4c9.png)
. By CRT, there must exist consecutive integers 
that are divisible by 
respectively, where 
is the 
th prime. Consequently, each 
is divisible by , and each 
is 
mod . Therefore, the 
, reduced mod , span the set {
}, so (exactly) one 
is divisible by .
, so in fact there are infinitely many satisfying your condition.
is just . By CRT, there must exist consecutive integers that are divisible by respectively, where is the th prime. Consequently, each is divisible by , and each is mod . Therefore, the , reduced mod , span the set {}, so (exactly) one is divisible by ., so in fact there are infinitely many satisfying your condition. should not only be a multiple of 
, but actually take the form of 
, where 
. This is something to fix - by specifying the remainder w.r.t 
. But CRT cannot apply directly. I'm still working on this. What do you think?
Prove that
Let 
Prove that
Let 
Prove that
such that 