If $x=123456789101112131415161718$, then $x\equiv 6\pmod{16}$ and $x\equiv 0\pmod 6$ BdMO 2013 Rajshahi

Build a number by writing down consecutive natural numbers starting from $1$ which  is 
  divisible by $6$ and gives a reminder of $6$ upon division by $16$. Such a number is $123456789101112131415161718$. To find the next such number up to which 
  number will you have to write?

For a number to be divisible by $6$,it must be divisible by both $2$ and $3$. Therefore,the number we are going to count up to must be even.Also,if any number divisible by 16,it's last 4 numbers must be divisible by 16..I tried building some equations but they yielded nothing.
Note that subtracting 6 will create a number that is divisible by both 6 and 16.I am sure we are going to use this somehow.A little hint will be appreciated.
 A: Hint $\ $ Suppose that $\ y\, =\, x19202122\cdots (n\!-\!1)n\,$ is the next one, where $\,19 \le n \le 99.$
Note $\,\ {\rm mod}\,\ 16\!:\,\ \color{#0a0}{10^2\equiv 4}\ \Rightarrow\,\color{blue}{10^4\equiv 0}\,$
so $\: 6 \equiv y\equiv n + \color{#0a0}{10^2}(n\!-\!1)+\color{blue}{10^4}((n\!-\!3)+\, \cdots\,)\, \equiv\, 5n\!-\!4\iff 5(n\!-\!2)\equiv0\overset{\!\!\!\!\!(5,16)\,\equiv1\!\!\!\!\!}\iff\! \color{#c00}{n\equiv 2}$
${\rm mod}\ 3\!:\,\ x,\,x19,\, x1920,\ldots$ $\smash{\overset{\!\!\!\!\!\!10^n\,\equiv\, 1\!\!\!}\equiv}\,\ x, x\!+\!19, x\!+\!19\!+\!20,\ldots \equiv\, 0,1,0,0,1,0,\ldots\equiv 0\iff \color{purple}{n\not\equiv 1}$  
Therefore we seek the least $\,\color{#c00}{n =2}+16k \in [19,99]\,$ such that $\,\color{purple}{n\not\equiv 1}\pmod{\! 3},\,$ which is $\,\ldots$
A: This eventually boils down to discovering all numbers $5a+d=8k$, where $a\in\{0,\ldots,9\}$ and $d\in\{0,\ldots,4\}$, then computing $b=2d+1$, and checking to see if the sum of digits of the number $n=\overline{123\ldots a(b-1)ab}\quad$ is a multiple of $3$.
