Inspired by Rama Devi

The $r^{th}$ term of the series is:

$T_r = 10(6)(10^0 + 10^1 + ... 10^{r-1})$

$T_r = 60\displaystyle \sum_{i=1}^{r} 10^{i-1}$

$T_r = 60 \dfrac{1(10^r - 1)}{9} = \dfrac{20(10^r - 1)}{3}$

The sum to n terms of the series is:

$S = \displaystyle \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} \dfrac{20}{3} 10^r - \sum_{r=1}^{n} \dfrac{20}{3}$

$\Rightarrow S = \dfrac{20}{3} \dfrac{10(10^n - 1)}{9} - \dfrac{20n}{3}$

$\Rightarrow S = \dfrac{20}{27} (10^{n+1} - 10 - 9n)$

$e = 20 , a = 27 , b = 10 , c = 9 , d = 10 \Rightarrow a + b + c + d + e = \boxed{76}$

$60(1+11+111+\dots)$

$\dfrac{60}{9}(9+99+999+\dots)$

$\dfrac{20}{3}((10-1)+(100-1)+(1000-1)+\dots)$

$\dfrac{20}{3}(10+100+1000+\dots+(10)^{n}-n$

$\dfrac{20}{3}(10(\dfrac{(10)^{n}-1}{10-1})-n$

$\dfrac{20}{27}((10)^{n+1}-10-9n)$

comparing from the above mentioned form, we get $a=27,b=10, c=9, d=10, e=20$

So, $a+b+c+d+e=\color{#D61F06}{\boxed{76}}$