Problem: Show that if 100 positive integers are selected from 1, 2,…, 200 such that no any two different numbers in the selection with one divisible by the other one, then no any one of 1, 2, …, 15 should be selected.
The problem is the same as
If 100 integers are selected from 1, 2, …, 200, and one selected number is from 1,2,…,15,
then there are two selected numbers with one divisible by the other.
Proof:
Let 
be the positive integers selected from 1, 2,
…, 200 such that no any two different numbers with one divisible by the
other one.
Presenting each number 
as 
where 
is the highest power of 2 as factor of 
and 
is the odd factor of.
Let
, then no any two different
are the same number in
.
Therefore, should contain 100 different odd numbers from 1 to 200.
.
Let us prove that set 
should not have a number from 1,2,…,15.
(1) Assume
that
, 
since 
. Therefore, there
exit two numbers,
, from A such that 
is divisible by 15. Contradiction.
(2) The
same argument, 3, 5, 7, 11, 13 should not be in .
(3) Assume
that 
,
Let
be from, then 
should not be the same positive integer.
(Otherwise, 
is divisible by 
). 
, otherwise is over 200. Therefore,
. There are two
numbers 
from A such that is divisible by 14.
(4) Using
from A and the same argument from
(3), we can prove that
.
(5) Using
from A and the same argument from (3), we can prove that 
.
(6) Consider
, since 
and 
. Therefore, 2 and 4
should not be in A.
(7) 1 should not be in A.
(8) If 9
is in A, consider 
. We have 
is divisible by 9.
Contradiction.
(9) If 12 is in A, then
Since
, 
with
because 12 is in A.
Since
, 
with 
because 12 is in A.
Since
, 
with 
because 12 is in A.
and
are the same number, otherwise contradict with the
assumption.
and
are the same number, otherwise contradict with the
assumption.
, otherwise 
. Therefore, 
, there are two numbers 2*45 and 2*135 in A. Contradiction.
(10) If 8 is in A, then 
could not be a factor for any number in A, unless
.
Therefore, 1, or 6, or 4 should be in A. Contradiction.