Solution 1
\[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\]
Solving by the quadratic formula, $n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19$ (since clearly $n \geq 0$). The answer is therefore $1 + 9 = \boxed{\textbf{(C) }10}$.

Solution 2
Dividing both sides by $n!$ gives
\[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.\]
Since $n$ is non-negative, $n=19$. The answer is $1 + 9 = \boxed{\textbf{(C) }10}$.

Solution 3
Dividing both sides by $n!$ as before gives $(n+1)+(n+1)(n+2)=440$. Now factor out $(n+1)$, giving $(n+1)(n+3)=440$. By considering the prime factorization of $440$, a bit of experimentation gives us $n+1=20$ and $n+3=22$, so $n=19$, so the answer is $1 + 9 = \boxed{\textbf{(C) }10}$.

Solution 4
Since $(n+1)! + (n+2)! = (n+1)n! + (n+2)(n+1)n! = 440 \cdot n!$, the result can be factored into $(n+1)(n+3)n!=440 \cdot n!$ and divided by $n!$ on both sides to get $(n+1)(n+3)=440$. From there, it is easier to complete the square with the quadratic $(n+1)(n+3) = n^2 + 4n + 3$, so $n^2+4n+4=441 \Rightarrow (n+2)^2=441$. Solving for $n$ results in $n=19,-23$, and since $n>0$, $n=19$ and the answer is $1 + 9 = \boxed{\textbf{(C) }10}$.
~Randomlygenerated

Solution 5
Rewrite $(n+1)! + (n+2)! = 440 \cdot n!$ as $(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!.$ Factoring out the $n!$ we get $n!(n + 1 + (n+1)(n+2)) = 440 \cdot n!.$ Expand this to get $n!(n^2 + 4n + 3) = 440 \cdot n!.$ Factor this and divide by $n!$ to get $(n + 1)(n + 3) = 440.$ If we take the prime factorization of $440$ we see that it is $2^3 * 5 * 11.$ Intuitively, we can find that $n + 1 = 20$ and $n + 3 = 22.$ Therefore, $n = 19.$ Since the problem asks for the sum of the didgits of $n$, we finally calculate $1 + 9 = 10$ and get answer choice $\boxed{\textbf{(C) }10}$.
~pnacham
There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$. What is the sum of the digits of $n$?
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$