are complex numbers with absolute value 
.
define 
.
converges. Show that,
.
such that there exists an integer polynomial 
with no integer roots and a constant 
such that
has at least 
distinct prime divisors for all positive integers 
.
is an integer, then there exist invertible matrices 
with non-zero entries such that:![\[A_1^{-1} + A_2^{-1} + \ldots + A_n^{-1} = (A_1 + A_2 + \ldots + A_n)^{-1}.\]](http://latex.artofproblemsolving.com/a/f/9/af9ff29771952abe37a00f3830c72e8c517e1cd6.png)
topic in College Math :coolspeak:
be a closed and countable set. Prove that if the analytic function 
is bounded, then 
is constant.
Over all real polynomials 
of degree 
what is the largest possible number of negative coefficients of 
fibonacci numbers 
)
and 
![$f:[1,8]\to R$](http://latex.artofproblemsolving.com/c/e/c/cec97ea0372e37977e8a77e8e77518a2344c1c43.png)
that:
Y by wu2481632, Davi-8191, samrocksnature, HWenslawski, math31415926535, jhu08, Adventure10, Mango247, ItsBesi, cubres, NicoN9
Find, with proof, the least integer 
such that if any 
elements are removed from the set 
, one can still find distinct numbers among the remaining elements with sum .
such that if any 
elements are removed from the set 
, one can still find distinct numbers among the remaining elements with sum .This post has been edited 1 time. Last edited by NormanWho, Apr 20, 2016, 9:31 PM
.
basically match up pairs to make 
and gg.
.
.![\[(1,6048), (2,6047), \ldots, (3024,3025),\]](http://latex.artofproblemsolving.com/3/c/6/3c605115e0b235fc3d163eeca788e427f14c7c3e.png)
.
(hopefully I typed that right, it's supposed to be 
)![\[2017+2018+\dots+4032=1008*6049,\]](http://latex.artofproblemsolving.com/2/f/5/2f54abf69dc0c24431ebcb670c5a4184c60904ab.png)
which we need when we remove the positive integers from 
,
,
pairs summing to 
,
,
.
full pairs left. If we take all of the numbers from any 
numbers and are looking for 
,
to 
.
.
from the set and this is the smallest sum we can get from the remaining elements. Now we prove this satisfies the condition. We note that there are 
.
such pairs left in the set, which allows us to sum 
,
.
as always possible 
N = 1008.6049.
.
.
.
.
.
.
,
which is 
and so we are done..
are removed than the least sum is 
Just like we wanted.
works. To prove this consider the pairs: 
Regardless of which 
are deleted, at least 
of these pairs have both elements remaining. Since each pair has sum 6049, we can take these pairs to be the desired numbers.
.
clearly follows from removing 
.
works, take the pairs 
.
remaining numbers summing to 
,
doesn't work. 
.
and 
that sum to 
By removing the first 
we see that 
Now, we prove validaty. If we don't remove any of 
we have that this will always work. Now, assume that we remove exactly 
is valid. For each removed 
within our range, consider removing 
from our sum, and replacing with a valid 
pair. Thus, we are done.
.
,
.
works. Consider the pairs 
.
.
,
,
