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

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

Video Solution 1

https://youtu.be/7p_ESPPF2es

~Education, the Study of Everything

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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