More random combo
by math154, Oct 10, 2011, 10:01 PM
1. (KöMaL) Let
be an integer and
be integers that give at least
distinct remainders when divided by
. Prove that some of these
numbers add up to a multiple of
.
Solution
2. (AMM) Let
be some points on the unit circle. Also let
be a regular polygon inscribed on this circle. Fix an integer
with
. Prove that one can find
such that
.
Solution






Solution
Assume for the sake of contradiction that no nonempty zero-sum subset exists.
To use pigeonhole modulo
(which is really the only reasonable way), we need to get both
and
involved. The easiest way to do this is to WLOG let
be pairwise distinct (we know such numbers exist by the problem hypothesis) and consider the partial sums
(which are pairwise distinct and nonzero by our assumption; we start at the
partial sum so that when we subtract elements from
we still have a sum of distinct
).
Now we need another set of at least
nonzero elements for pigeonhole to work; something like
quickly comes to mind (since the
are distinct, at most one of them can satisfy
, and we can afford to lose a few), provided we make it compatible with
. But this is easy, since adding
to every entry to get
does not make it less compatible with
.
WLOG
for
(we can assign the roles to
and
if necessary); then the sets
,
, and
each contain pairwise distinct nonzero elements modulo
. But they have
elements in all, so by pigeonhole some two sets must intersect (we are only dealing with the
nonzero elements). If the third set intersects with either the first or the second, we obviously get a nonempty zero-sum subset; otherwise, if the first two sets intersect, then we must have
and
for some
, which also gets a nonempty zero-sum subset since
contains at least
terms.
To use pigeonhole modulo








Now we need another set of at least








WLOG















2. (AMM) Let






Solution
For a real number
, let
denote how far
is from its closest integer multiple of
(i.e. absolute value
). If we go by contradiction, then we have that
, where
represent the positions of
on an angular scale. To utilize this local information (which restricts what points can be close to each other) in the context of our global situation of
reals spread out
, we want to get a lot of the
in some interval
.
On average,
contains
of the
. If some
contains
or more of the
, then two of the indices
must satisfy
, contradiction. Otherwise, the interval
contains exactly
of the
for almost all
(except for a set of zero density), which is impossible as we consider
for any
and sufficiently small
unless
for some
; but then we get
or more
in
.











![$[r,r+k-1]\pmod{n}$](http://latex.artofproblemsolving.com/4/4/4/44438de8f829bc08c1f7cc2c7d8c62b6bdc32a7b.png)
On average,
![$[r,r+k-1]\pmod{n}$](http://latex.artofproblemsolving.com/4/4/4/44438de8f829bc08c1f7cc2c7d8c62b6bdc32a7b.png)


![$[r,r+k-1]\pmod{n}$](http://latex.artofproblemsolving.com/4/4/4/44438de8f829bc08c1f7cc2c7d8c62b6bdc32a7b.png)




![$[r,r+k-1]\pmod{n}$](http://latex.artofproblemsolving.com/4/4/4/44438de8f829bc08c1f7cc2c7d8c62b6bdc32a7b.png)



![$r\in[p_m-\epsilon,p_m+\epsilon]$](http://latex.artofproblemsolving.com/c/e/e/ceee1e2b307b90066a1f13d56ed4ec3506c10b2d.png)






![$[p_m,p_m+k-1]\pmod{n}$](http://latex.artofproblemsolving.com/8/7/0/87004e87cb45a57ee073a86a5d703d1d750bfbf0.png)
This post has been edited 5 times. Last edited by math154, Oct 11, 2011, 2:01 PM