smc

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