TITLE: Choose 100 numbers from 1~200 (one less than 16) - prove one is divisible by another!
QUESTION [19 upvotes]: Prove that if 100 numbers are chosen from the first 200 natural
  numbers and include a number less than 16, then one of them is
  divisible by another.

How to prove this? many thanks....

REPLY [2 votes]: Let us suppose that we have chosen 100 positive integers , not exceeding 200 ,none of which is divisible by any other. Let us prove that none if the numbers from 1 to 15 is contained among these 100 numbers.
Let us consider all the greatest odd divisors of the chosen numbers .It is obvious that these divisors from the set of all odd numbers not exceeding 200 .In particular ,these odd divisors there are no two numbers one of which is divisible by the other,the number containing the odd factor 27 must be divisible by a power of 2 whose exponent is at least one ,the number containing the odd factor 9 must be divisible by a power of 2.Whose exponent is atleast 2,the number containing the odd factor 3 must be divisible by a power of 2 whose exponent is atleast 3,the number containing the odd factor 1must he divisible by a power of 2 whose exponent is atleast 4.This means that the number 1,2=1.2, 2,3,4=1.2^2, 6=2.3, 8=1.2^3,  9 and 12 =3.2^2 are not contained among the 100 choen numbers.
In just the same way we can consider those of the given numbers whose greatest odd divisors are 5,15 and 45 and prove the given numbers do not donation, 5,10=5.2 and 15.Similarly , if we consider the number 7 ,21 and 63 ,we can show that the numbers 7 and 14 are not among the 100 chosen integers; while if ee consider the numbers 11 and 33 ,we can show that 11 is not amobg the 100 chosen integers and if we consider the numbers 13and 39,we can show that 13 is not among the 100 chosen integers.