Inspired by Sakanksha Deo

Here is a tricky part in this problem. Let's investigate that. $6+66+666 + \ldots$ We will try to convert it in the form so that we may create a $GP$ in it. $\dfrac{2}{3}(9+99+999+ \ldots)$ $\dfrac{2}{3}(10-1+10^{2}-1+10^{3} - 1 + \ldots)$ Now we have created a $GP$ . Now it can be summed easily using formula for finite $GP$ summation. $\dfrac{2}{3}(\sum_{r=1}^{n}( 10^{r} ) - n)$ $\dfrac{2}{3} (\dfrac{10(10^{n}-1}{10-1}) - n)$ Now we just have to manipulate the expression to get the expression in desired format. $\dfrac{2}{3}\dfrac{10^{n+1} - 9n-10}{9}$ $\dfrac{2}{27}(10^{n+1} - 9n-10)$

Therefore $27+10+9+10+2=\boxed{58}$

$6+66+666+6666+...=6(1+11+111+1111+...)$

Partial sums of the part in the parentheses: $\{1,12,123,1234...\}$

Formula for the $n$ th partial sum via a recurrence relation:

$a(n)=10a(n-1)+n$

This equation is inhomogeneous. Let's turn it into a homogenous recurrence relation and solve by substituting $a(n)=r^x$ :

$a(n)=10a(n-1)+n$ , $a(n-1)=10a(n-2)+n-1$

$a(n)-a(n-1)=(10a(n-1)+n)-(10a(n-2)+n-1)$

$a(n)=11a(n-1)-10a(n-2)+1$ , $a(n-1)=11a(n-2)-10a(n-3)+1$

$a(n)-a(n-1)=(11a(n-1)-10a(n-2)+1)-(11a(n-2)-10a(n-3)+1)$

$a(n)=12a(n-1)-21a(n-2)+10a(n-3)$ , replace $a(n)=r^x$

$r^x=12r^{x-1}-21r^{x-2}+10r^{x-3}$

$r^3-12r^2-21r+10=0$

$(r-10)(r-1)^2=0\Longrightarrow\{1,10\}$

We now must take our values of $r$ and use them to create a general formula for $a(n)$ :

$a(n)=c_110^n+c_2x1^n+c_31^n=c_110^n+c_2n+c_3$

Knowing that $a(1)=1$ , $a(2)=12$ , and $a(3)=123$ , let's solve for $c_1$ , $c_2$ , and $c_3$ :

$10c_1+c_2+c_3=1$

$100c_1+2c_2+c_3=12$

$1000c_1+3c_2+c_3=123$

$90c_1+c_2=11$

$900c_1+c_2=111$

$810c_1=100\Longrightarrow c_1=\frac{10}{81}$

$90\left(\frac{10}{81}\right)+c_2=11\Longrightarrow c_2=-\frac{1}{9}$

$10\left(\frac{10}{81}\right)-\frac{1}{9}+c_3=1\Longrightarrow c_3=-\frac{10}{81}$

Plugging into our equation:

$a(n)=\frac{10}{81}10^n-\frac{9}{81}n-\frac{10}{81}=\frac{1}{81}(10^{n+1}-9n-10)$

Now that we have our formula for the partial sums of the part within the parentheses, our answer is simply $6$ times $a(n)$ :

$6a(n)=\frac{6}{81}(10^{n+1}-9n-10)=\frac{2}{27}(10^{n+1}-9n-10)$

Finally, we get $a=27$ , $b=10$ , $c=9$ , $d=10$ , and $e=2$ . Therefore, our answer must be:

$27+10+9+10+2=\boxed{58}$