Sum of digits equality $Z$ return sum of digits of number: $Z(15)=6$.
If for some $W\in\mathbb{N}$, $Z(W)=100;~Z(44W)=800$ find
$$
ZZ(2015!)+ZZZ(2015!)+ZZZZ(2015!)+Z(3W)
$$
I don't have experience with this kind of problem, please give a hint what should I start? What approach is right?

The sum of digits function returns about $4.5$ times the base $10$ log of a number, because the base $10$ log gives the number of digits and the average digit is $4.5$.  It also maintains the value of the number $\bmod 9$ by the classic divisibility test.  $2015!$ has about $5500$ digits (you can use Stirling to get the exact answer), so $Z(2015!) \lt 5500 \cdot 9 =49500$  Then $ZZ(2015!) \lt 4+8+9+9+9=49$ and since $2015!$ is a multiple of $9$, we have $ZZ(2015!)=9, 18, 27$ or $36$ and you should be able to get the next two terms of your sum.  I don't see an easy way to distinguish the four cases for the first term.  For the last term, multiplying $W$ by $44$ multiplied $Z(W)$ by $8=Z(44)$ so you can't afford any carries of one non-zero digit into another.  What does that tell you about $Z(3W)$?
