high-quality questions. This test was created by multiple 130+ scorers, and we have tried our very best to produce the highest quality test possible. We've blended the question types in a good distribution, and maintained a realistic difficulty. (something most AOPS mocks fail at)
such that![\[f(x)+f(t)=f(y)+f(z)\]](http://latex.artofproblemsolving.com/7/e/c/7ecb395c1e56e6d9726fc45f1bc7a66d98333da2.png)
for all rational numbers 
that form an arithmetic progression. 
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.
,
pairs have sum 
)
complete pairs of 
.
forces 
,
.
,
,
is achievable. There are 
.
.![\[(1,6048), (2,6047), \ldots, (3024,3025),\]](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 
to 
,
,
,
,
.
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 
of those. Then, we have that there are at least 
is valid. For each removed 
within our range, consider removing 
from our sum, and replacing with a valid 
pair. Thus, we are done.
