Sum of digits

Every number is congruent mod $9$ to its digit sum. We can compute the factors of $N$ mod $9$ by repeatedly iterating digit sums. So $\begin{aligned} N &\equiv 4 \cdot 8 \cdot 5 \cdot 19 \cdot 53 \mod 9 \\ &\equiv 4 \cdot 8 \cdot 5 \cdot 1 \cdot 8 \mod 9 \\ &\equiv 2 \mod 9 \end{aligned}$ If $k$ is the digit left out of $N$ , then the digit sum of $N$ is $90-2k$ . This is $2$ mod $9$ , so we get $k \equiv 8$ mod $9$ , so $k = 8$ and the digit sum of $N$ is $\fbox{74}$ .