Crazy functions!!!!

We must first try to find the smallest possible value of $N$ . By doing so, we must put in as many $9$ 's as possible. Since $\left \lfloor \dfrac{2013}{9} \right \rfloor=223$ , there are $223$ nines and one $6$ in this number $N$ (the last number is a 6 because $2013-9 \cdot 223=6$ . To make this the smallest positive integer, the $6$ must be the leftmost digit. From here, we must multiply by $5$ . We can dissect this massive number as $5 \cdot (6\underbrace{000 \ldots 00}_{223 \text{ zeroes}} + \underbrace{9999 \ldots 999}_{223 \text{ nines}} )$ . The first part gives us $3\underbrace{000 \ldots 00}_{224 \text{ zeroes}}$ , while the second part gives us $4\underbrace{999 \ldots 99}_{222 \text{ nines}}5$ . Adding them up gives us $34\underbrace{999...99}_{222 \text{ nines}}5$ which is equivalent to $35\underbrace{000 \ldots 00}_{223 \text{ zeroes}}-5$ Since our objective is to find $S(5N+2013)$ , we must add the number above to $2013$ and then find the digit sum. $35\underbrace{000 \ldots 00}_{223 \text{ zeroes}}-5+2013=35\underbrace{000 \ldots 00}_{223 \text{ zeroes}}+2008=35\underbrace{000 \ldots 00}_{219 \text{ zeroes}}2008$ Since we conveniently do not need to add any of the zeroes, our final answer is $3+5+2+8=\boxed{18}$

Any questions are welcome, and I will do my best to answer them.

N is least integer satisfying given condition so N = 6followed by 223 9s Using this calculate required value