Is there a number in the form 201820182018...2018
with 2018 repeating n times that is a multiple of 2019?
0.111...*9=10.222...*9=2
0.2525252...*99=25
0.2018201820...*9999=2018
0.2018201820...*99999999=20182018
0.2018201820...* (9999... n ....9999)=(2018... n ... 2018)
Call p=(2018... n ... 2018)
2019=673*3 (no other divisors) => p will be multiple of 2019 if it is multiple of 673 and is multiple of 3
Let see when p is multiple of number prime 673
1/673=0,001485884... as a notregular fraction the decimal part is repeated after 672 digits or less. Concretely for number 673 the decimal part has a period of 224 digits. (Big online numbers calculator for this)
As we have seen in the beginning of the post:
0,001485884... (follows digits since 224, and repeated infinite times) can be express as:
001485884.../9999...224 nines...999 (where both numbers are integers).
Then I can choose number p (=2018... n ....2018) with n=56 groups (2018) and with a number of digits 56*4=224 digits.
p/673=p*0,01485884...
=p*(001485884.../9999...56 groups..9999)
=p*001485884...*(0,20282028.../p)
So:
p=673*001485884...*0,20282028....
This is an integer as p is an integer an, we can see it, is a multiple of 673. In addition 673 is prime so 001485884...*0,20282028.... is an integer.
Let see when p is not just multiple of 673 but of 2019 (673*3).
Here we will need to work with three full decimal periods of 1/673.
(0.001485884....001485884....001485884)...001485...=
=001485... 001485... 001485 integer with 224*3 digits *
* 9999... 224*3 nines ...9999
See that now: =001485... 001485... 001485 integer with 224*3 digits is obviously a multiple of 3.
Repeating what we did before, the new p composed of groups of number 2018 will be:
p=673*001485... 001485... 001485...*0,20282028...
The product of the last two terms will be an integer again. Then:
p will be now multiple of 2019, will have now 224*3 digits=672 digits, so that it will be composed by 168 groups of 2018.
Edited on March 4, 2019, 3:04 pm

Posted by armando
on 20190304 06:30:07 