In: Math
Let S be the set of natural numbers which can be written as a non-empty string of ones followed by a non-empty string of zeroes. For example, 10, 111100 and 11100000 are all in S, but 11 and 1110011 are not in S. Prove that there exists a natural number n∈S, such that 2018 | n.
Consider the set 

{ where the last number in the set contains 2019 such 
 }.
There are 2019 numbers in the set 
. Now, we consider their remainders when divided by 
 . See that when any number is divided by 
 , the possible remainders are 
 {i.e. there are 2018 possible remainders}. Since 
 has 2019 numbers, by the Pigeonhole Principle at least two of the
numbers in 
 give the same remainder when divided by 
 .
Let those numbers be 
 and 
 . Then we have 

i.e. when 
 divides 
 then the remainder is 
 .
i.e. 
    { where 
 }.
Note that the difference between any two numbers in the set
 is of the form 
{ i.e. a string of 
 followed by a string of 
 , which are precisely the elements of 
 }.
So 
 . Put 
 . Then we have an 
 such that 
 .
This completes our proof.