Difference between revisions of "1992 USAMO Problems/Problem 3"
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− | Furthermore, if you add any amount of elements to the end of a <math>n</math>-<math>p</math> bad set to form another <math>n</math>-<math>p</math> set (with a different <math>n</math>), it will stay as a <math>n</math>-<math>p</math> bad set because <math>a_{n+x}>a_{n}>p+1</math> for any positive integer <math>x</math> and <math>\ | + | Furthermore, if you add any amount of elements to the end of a <math>n</math>-<math>p</math> bad set to form another <math>n</math>-<math>p</math> set (with a different <math>n</math>), it will stay as a <math>n</math>-<math>p</math> bad set because <math>a_{n+x}>a_{n}>p+1</math> for any positive integer <math>x</math> and <math>\nexists \sigma(Y)=p+1</math>. |
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<math>\{1,2,4,8,16,32,64,128,247,248,750\}</math> | <math>\{1,2,4,8,16,32,64,128,247,248,750\}</math> | ||
− | Note that the first 8 numbers are power of <math>2</math> from <math>0</math> to <math>7</math>, and realize that any <math>8</math> or less digit binary number is basically sum of a combination of the first <math>8</math> elements in the set. Thus, <math>\ | + | Note that the first 8 numbers are power of <math>2</math> from <math>0</math> to <math>7</math>, and realize that any <math>8</math> or less digit binary number is basically sum of a combination of the first <math>8</math> elements in the set. Thus, <math>\exists Y\subseteq\{1,2,4,8,16,32,64,128\}</math>, <math>\sigma(Y)=x \forall 1\le x\leq 255</math>. |
− | <math>248\le\sigma(y)+a_9\le502</math> which implies that <math>\ | + | <math>248\le\sigma(y)+a_9\le502</math> which implies that <math>\exists A\subseteq\{1,2,4,8,16,32,64,128,247\}</math>, <math>\sigma(A)=x \forall 1\le x\leq 502</math>. |
− | + | Similarly <math>\exists B\subseteq\{1,2,4,8,16,32,64,128,247,248\}</math>, <math>\sigma(A)=x \forall 1\le x\le750</math> and <math>\exists C\subseteq\{1,2,4,8,16,32,64,128,247,248,750\}</math>, <math>\sigma(A)=x \forall 1\leq x\leq 1500</math>. | |
<br/>Thus, <math>\{1,2,4,8,16,32,64,128,247,248,750\}</math> is a <math>11</math>-<math>1500</math> set. | <br/>Thus, <math>\{1,2,4,8,16,32,64,128,247,248,750\}</math> is a <math>11</math>-<math>1500</math> set. | ||
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Done but not elegant... | Done but not elegant... | ||
− | == | + | == See Also == |
{{USAMO box|year=1992|num-b=2|num-a=4}} | {{USAMO box|year=1992|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Olympiad Combinatorics Problems]] |
Latest revision as of 07:52, 19 July 2016
Problem
For a nonempty set of integers, let
be the sum of the elements of
. Suppose that
is a set of positive integers with
and that, for each positive integer
, there is a subset
of
for which
. What is the smallest possible value of
?
Solution
Let's a -
set be a set
such that
, where
,
,
, and for each
,
,
,
,
.
(For Example is a
-
set and
is a
-
set)
Furthermore, let call a -
set a
-
good set if
, and a
-
bad set if
(note that
for any
-
set. Thus, we can ignore the case where
).
Furthermore, if you add any amount of elements to the end of a -
bad set to form another
-
set (with a different
), it will stay as a
-
bad set because
for any positive integer
and
.
Lemma) If is a
-
set,
.
For ,
or
because
and
.
Assume that the lemma is true for some , then
is not expressible with the
-
set. Thus, when we add an element to the end to from a
-
set,
must be
if we want
because we need a way to express
. Since
is not expressible by the first
elements,
is not expressible by these
elements. Thus, the new set is a
-
set, where
Lemma Proven
The answer to this question is .
The following set is a -
set:
Note that the first 8 numbers are power of from
to
, and realize that any
or less digit binary number is basically sum of a combination of the first
elements in the set. Thus,
,
.
which implies that
,
.
Similarly ,
and
,
.
Thus, is a
-
set.
Now, let's assume for contradiction that such that
is a
-
set where
is a
-
set where
(lemma).
Let be a
-
set where the first
elements are the same as the previous set. Then,
is not expressible as
. Thus,
.
In order to create a -
set with
and the first
elements being the ones on the previous set,
because we need to make
expressible as
. Note that
is not expressible, thus
.
Done but not elegant...
See Also
1992 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.