When you add the digit 1 at the end of this number you get a number thrice the number with 1 added before it.
Evaluate this number.
To restate the problem, N1 = 3 * 1N (where 1 and N are concatenated).
Or, 10x + 1 = 3*(10^n + x) = 3*10^n + 3x
7x = 3*10^n  1
so 7x is of the form 2, 29, 299, 2999, etc. This series, mod 7 goes through a cycle of 6, every 6th one being 0 mod 7.
The first of these is 299999 which divided by 7 is:
42857.
And 428571 / 142857 = 3
So, solutions include:
42857, 42857142857, 42857142857142857, ...
In the limit, our original concatenations, N1 / 1N (if you put a decimal point in front of them) become 3/7 divided by 1/7

Posted by Larry
on 20180102 08:31:51 