Show that there exists an integer all of whose digits are on

Show that there exists an integer all of whose digits are ones such that it is divisible by 2017. Not 2017 is divisible by a number. (2017 is prime)

Solution

111....111 - 11111....111 = 1111....11000....000 = 1111...1*10m

gcd(2017,10m)=1

So the statement is equivalent to showing 2017|1111....1 for some 1111....1.

2017 is prime.

So it suffices to show p|111....1 for some 1111..11 for each prime p

.10p11modp by Fermat\'s little theorem. So $p|10^{p-1} - 1.

Now (10^(p1)1)/(101)=j=0 to (p-2) 10^i = 1111....1 with p1 1s.

So p| 9 x (10^(p1)1) /(101)

. If p3 the gcd(p,9)=1 and p|10^(p1)1/(101). (If p=3 then 3|111)

So 2017|(10^20161)/9 which is 2016 1s.

So.... N = 111111111...1111 with with k>2016

1s and M = 11111....111 with k2016 1s. is such that N-M = 111....1111000.....0 which has 2016 1s and k-2016 0s. This is divisible by 1111....1111 with 2016 0s which is divisible by 2017.

Hence, there exists an integer which is divisible by 2017.