# Difference between revisions of "2019 AMC 10B Problems/Problem 6"

The following problem is from both the 2019 AMC 10B #6 and 2019 AMC 12B #4, so both problems redirect to this page.

## Problem

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

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

## Solution

### 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 $1 + 9 = \boxed{\textbf{(C) }10}$.

~ pi_is_3.14

~IceMatrix

~savannahsolver