AoPS Academy
with the aim to train for the first problem of . They play the following game.
bacteria and choose a natural number 
. On each move, the player chooses any 
number from to . Then the player divides each bacterium into pants. Once chosen, the number cannot be chosen twice. If after any player's move the number of bacteria population is divisible by then that player loses. Determine who has the winning strategy depending on the given number if it's known that Amman starts first.
where 
is the inverse gamma function, and 
is the inverse of 
and 
then is this uniformly convergence on 
comment on uniformly convergence on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
where in general it is should be uniformly convergence.
and trying to solve
![$f:[-1,1] \rightarrow \mathbb{R}$](http://latex.artofproblemsolving.com/d/d/4/dd4186fc296a916a1aedeb1f3b59ce20034e99d3.png)
be a continuous function such that 
. Prove that
![\[I=\int\limits_{-\infty}^{\infty}e^{x-10\cosh (2x)}\mathrm dx.\]](http://latex.artofproblemsolving.com/5/b/6/5b61a79360c164fcd935eb613e12ab056847a93c.png)
be a linear transformation on linear space 
satisfying:
but 
, and 
is the eigensubspace of eigenvalue 
. Prove that can be decomposed to 
-cyclic subspace's direct sum. over the field 
:
and 
functions are given with following properties:
is strict increasing and for each 
there holds 
or 
.
if holds and 
if holds.
and 
such that for all 
we have 
and 
be nonnegative numbers and 
be a nonnegative symmetric matrix such that 
. Given 
, consider the quadratic form
Assume that
there exists 
such that 
-the entry of the 
-th power 
of 
is positive. Show that 
has a unique minimum and the minimum lies in the open cube 
in 
.
such that for all 
, we have![$$
\sin y f(x)-\sin x f(y) \leq \sqrt[2025]{(x-y)^{20226}}
$$](http://latex.artofproblemsolving.com/8/9/6/896d8a02b5694dd3b18a9dc36a59f9fbcd9737f2.png)
be a ring not assumed to have an identity, with the following properties: that is not nilpotent.
are nonzero elements of , then 
. is a division ring, that is, the nonzero elements of R form a group under multiplication.
that's manifestly true. Let 
be polynomial of degree 
, so
where 
is a monic real polynomial of degree and 
is a polynomial of degree with only real roots and real coefficients.
, where 
, so :
... (q.e.d)
![\[ P = b\prod^n_{i=1} (x-a_i)\]](http://latex.artofproblemsolving.com/4/f/f/4ff85858b0e206f5e31e0681d1cc6f1be3a8d9f7.png)
is not a root of 
and Let 
, then![\[ P' = P (c_1 + c_2 + \ldots + c_n), P'' = P (\sum_{i \neq j} 2c_i c_j)\]](http://latex.artofproblemsolving.com/0/3/1/0314153b422f93f1673a5b74fc59c59a3f2a82b3.png)
![\[ (n-1)(c_1 + c_2 + \ldots + c_n)^2 \ge 2n (\sum_{i \neq j}c_i c_j)\]](http://latex.artofproblemsolving.com/6/7/8/67818283bcdf49b5a6ba5660006db786af8a0c55.png)
![\[ \sum_{i \neq j} (c_i - c_j)^2 \ge 0\]](http://latex.artofproblemsolving.com/5/b/a/5ba21af2e92ced15e6f6e0817da00bf21ba68f5a.png)
are equal, which means all the 
are equal and 
be the zeroes (non-necessarily distinct) of . The given inequality holds trivially for 
. So we assume that 
in the latter part of the solution.![$P'(x) = P(x) [\sum_{i=1}^n\frac{1}{x-\alpha_i}]$](http://latex.artofproblemsolving.com/4/f/8/4f845e30f7bc73200410de9b56f19dd85763f6b7.png)
![$$n\cdot\sum\limits_{i=1}^n\frac{1}{(x-\alpha_i)^2}\ge \left[ \sum\limits_{i=1}^n\frac{1}{|x-\alpha_i|}\right]^2\ge \left[\sum\limits_{i=1}^n\frac{1}{x-\alpha_i}\right]^2...(i)$$](http://latex.artofproblemsolving.com/7/0/7/70763c163e76fcfebb27909d534b3dcaf9420742.png)
can be re-written as ![$-n\left[\frac{P'(x)}{P(x)}\right]'\ge \left[\frac{P'(x)}{P(x)}\right]^2\iff 0\ge \frac{n(P''(x)P(x)-P'(x)^2)}{P(x)^2}+\frac{P'(x)^2}{P(x)^2}$](http://latex.artofproblemsolving.com/b/1/a/b1ae32c963595bc47ec0a1c85aca9f91e05a3ffe.png)
for all 
s are equal (equality case of Cauchy -Schewarz)
be the zeroes (non-necessarily distinct) of . The given inequality holds trivially for . So we assume that in the latter part of the solution. can be re-written as for all s are equal (equality case of Cauchy -Schewarz)
be the zeroes (non-necessarily distinct) of . The given inequality holds trivially for . So we assume that in the latter part of the solution. can be re-written as for all s are equal (equality case of Cauchy -Schewarz)
but you are excluding the roots.....this is the first problem so I can consider that 
but in that case if it's a constant polynomial then it has infinite roots...that is basically for all real ....then your statement is totally wrong....I don't think the 
idea is good for admitting all values of in real.....so if possible please post another solution....we are waiting for that
![\[I=\int\limits_{0}^{1}\frac{\ln x}{\sqrt{1-x^2}}\mathrm dx.\]](http://latex.artofproblemsolving.com/4/4/0/440e26c1aeee699f15a2564a5c9b63ccd94a4cb2.png)
.
