Is there a positive integer divisible by 2018 which only consists of the digits 3 and 0 (in base 10)?
As suggested by the title, a nonconstructive proof can be made using the Pigeonhole Principle.
Let S be the set of 2019 numbers which are repdigits of 3 up to 2019 digits. S={3, 33, 333, 3333, ......, 33....33 (2019 3's).
Divide each number in S by 2018 and note the remainders. By the Pigeonhole Principle at least two of the remainders are the same.
Then calculate N by taking the difference of two elements of S with the same remainder. N will consist of a series of 3's followed by a series of 0's and will be a multiple of 2018.
A constructive proof can be made by using Fermat's Little Theorem.
The prime factorization of 2018 is 2*1009. By Fermat's Little Theorem 10^(1008)-1 = 0 mod 1009. 10^(1008)-1 is a number consisting of 1008 9's and is a multiple of 1009.
10 is even and 3 is coprime to 1009. Then let N=10*(10^(1008)-1)/3. N is a 1009 digit number consisting of 1008 3's followed by a single 0, and N is a multiple of 1009*2=2018.
Solution
