Question 389515
a_1 = 111 = {{{111*10^0 = 111*((10^3 - 1)/(10^3 - 1))}}}
a_2 = 111 111 111 = {{{111*(((10^3)^3 - 1)/(10^3 - 1)) = 111*((10^(3^2) - 1)/(10^3 - 1))}}}
a_3 = 111 111 111...111  (27 1's) = {{{111*(((10^3)^(3^2) - 1)/(10^3 - 1)) = 111*((10^(3^3) - 1)/(10^3 - 1))}}}
........
a_n = {{{111*(((10^3)^(3^(n-1)) - 1)/(10^3 - 1)) = 111*((10^(3^n) - 1)/(10^3 - 1))}}} by induction.

Then{{{a_n/(3a_(n-1)) 
= 111*((10^(3^n) - 1)/999)/( 3*111*((10^(3^(n-1)) - 1)/999)))}}}, substituting 999 for {{{10^3 - 1}}}.
{{{a_n/(3a_(n-1))  = (10^(3^n) - 1)/(3*(10^(3^(n-1)) - 1)) = (10^(2*3^(n-1)) + 10^(3^(n-1)) + 1)/3}}}


Now {{{(10^(2*3^(n-1)) + 10^(3^(n-1)) + 1)/3}}} is definitely a positive integer, because the numerator is a positive integer with {{{2*3^(n-1)+1}}} digits, {{{2*3^(n-1) - 2}}} of which are 0's, and 3 are 1's.  Hence the sum of the digits of the numerator is 3, which makes the number itself divisible by 3.