2019 AMC 10B Problems/Problem 6

Revision as of 22:39, 14 February 2019 by Sulfurdioxideso2 (talk | contribs)
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 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 1

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

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

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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 positive, $n=19$. The answer is $1+9=10\Rightarrow \boxed{C}.$

Solution 3

Divide both sides by $n!$:


$(n+1)+(n+1)(n+2)=440$

factor out $(n+1)$:

$(n+1)*(n+3)=440$


prime factorization of $440$ and a bit of experimentation gives us $n+1=20$ and $n+3=22$, so $\boxed{n=19}$, so the answer is $\boxed{C}$

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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