Factorials are powerful!

A zero at the end of a number corresponds to having a pair of factors, a 2 and a 5. Since factors of two are always more plentiful than fives in a factorial (you can look at the first handful to convince yourself of this), we need only work out how many factors of 5 there are in 1005! Luckily, there’s a quick way to do this:

Each number less than 1005 which is divisible by five contributes a facor of five, so we’ll start with $\lfloor \frac{1005}{5} \rfloor = 201$

Each number divisible by 25 contributes two fives, but one of these 5’s has already been accounted for, so we only count one more: $\lfloor \frac{1005}{25} \rfloor = 40$

Similar reasoning gives $\lfloor \frac{1005}{125} \rfloor = 8$ extra factors and $\lfloor \frac{1005}{625} \rfloor = 1$ extra, for a total of 250 trailing zeros.

For some reason the question asks for the digital sum less four, so the answer is $\boxed 3$

(If you still aren’t convinced about the factors of 2, try adding them up the same way I’ve done with the 5’s. You’ll get a much larger answer, so we only need to know the number of fives.)