2006 Romanian NMO Problems/Grade 10/Problem 1
Contents
[hide]Problem
Let be a set composed of
elements and let
be its power set. Find all functions
that have the properties
(a) , for
;
(b) , for all
, where
.
Solution
Solution 1
I claim that , for all
. Clearly this function works; we must now show that it is the only function with the two given properties. We shall do this by proving that any such function
that satisfies both properties must be
.
We have , hence
. Also, if
, then
since
,
, and
.
Let be an increasing sequence of subsets of
, such that
for all
. Then
for all
.
Note that for any
, from property (a). Hence
which implies
for all
.
Solution 2
I claim that , for all
. Clearly this function works; we must now show that it is the only function with the two given properties. We shall do this by constructing a function
that does satisfy both properties, and then showing that this function is in fact
.
We have , hence
. Note that if
, then
, from property (a). Let
and
for some
. We then have from property
that
. Since
,
. This shows that for all
with
,
.
Lemma:for all
. Proof: We proceed by induction on
. Clearly
and
if
from property (a). This is our base case. Now we do the inductive step. Assume that for all
for some positive integer
,
if
. Let
and
be subsets of
such that
,
and
. We then have from property 2 that
. From our inductive hypothesis, this sum is at least
. This completes the inductive step, which completes the proof.
I shall now prove that for all
.
Note that our lemma implies that . Consider a set
with cardinality
. We then have that
and
. We then have that
and
, so their sum (which equals
from property (b)) is at least
. This then implies that
and
for all
. We defined
to be
, so
. This completes the proof.