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

Revision as of 22:42, 16 February 2019

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

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

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

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

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 $n=19$ , so the answer is $1 + 9 = \boxed{\textbf{(C) }10}$

Solution 4

Obviously n must be very close to $\sqrt{440}$ . By quick inspection, $n = 19$ works. $1 + 9 = \boxed{\textbf{(C) }10}$

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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 {{AMC12 box|year=2019|ab=B|num-b=3|num-a=5}}
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