AoPS Academy
+1 w
such that for every polynomial 
with integer coefficients and for every integer 
the integer![\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\]](http://latex.artofproblemsolving.com/a/2/7/a2793d7f46345bcba4ccaa3082cc47afcd35074b.png)
(the -th derivative of at 
) is divisible by 
has always integral solutions for 
and 
whenever 
is a positive integer.
![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
, there exists some constant 
, for any 
, among the sequence 
and 
could be chosen to satisfy 
and 
.
be a ring with unity in which 
and there is a morphism 
from the group 
to the monoid 
such that for all 
, there is a 
such that 
. Prove that is a field.
let the numbers 
be determined by the rules 
and 
Find the value of ![\[\sum_{n=1}^{2013}c(n)c(n+2).\]](http://latex.artofproblemsolving.com/3/f/a/3faf9da9b035e02298650c79251370d0c8b61c1f.png)
be expressed as a polynomial in 
with real coefficients?
be a twice differentiable function such that![\[
f(0) = 2, \quad f'(0) = -2, \quad \text{and} \quad f(1) = 1.
\]](http://latex.artofproblemsolving.com/1/e/1/1e178debf8ae0b6b76d2e60ffca95f5a4c4ad4e5.png)
Prove that there exists a point 
such that![\[
f(\xi) \cdot f'(\xi) + f''(\xi) = 0.
\]](http://latex.artofproblemsolving.com/4/6/b/46b8d103981b838ae0aa5afdee104edfae4e4b8f.png)
positive such that 
, then there exists 
positive such that 
converges and 
diverges ?
is continuous, positive with 
, does continuous and positive exist on 
such that 
converges and 
diverges ?
is in the 
class, increasing : I chose 
. With this idea, I had the idea to define 
but it is not clear that it is ok, even if is increasing. positive such that , then there exists positive such that converges and diverges ? And if is increasing ? positive increasing such that 
, then 
diverges ? positive increasing such that , then 
converges ? is positive increasing and such that 
does not converge to 
, can diverge ? only to be continuous ?
,
.
,![\[ \dfrac {1}{2ne} < \dfrac {1}{e} - \left( 1 - \dfrac {1}{n} \right)^n < \dfrac {1}{ne}. \]](http://latex.artofproblemsolving.com/1/3/e/13e2e3859e16ca0e4453330eeac03465d2eb8429.png)
