2019 AMC 10B Problems/Problem 6

There is a real $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$ ?

$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$

Solution

$(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!$ $n![n+1 + (n+2)(n+1)] = 440 \cdot n!$ $n + 1 + n^2 + 3n + 2 = 440$ $n^2 + 4n - 437 = 0$

$\frac{-4\pm \sqrt{16+437\cdot4}}{2} \Rightarrow \frac{-4\pm 42}{2}\Rightarrow \frac{38}{2} \Rightarrow 19$ . $1+9 = \boxed{C) 10}$

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