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Difference between revisions of "2019 AMC 10B Problems/Problem 6"
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<cmath>(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.</cmath>
 
<cmath>(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.</cmath>
 
Since <math>n</math> is positive, <math>n=19</math>. The answer is <math>1+9=10\Rightarrow \boxed{C}.</math>
 
Since <math>n</math> is positive, <math>n=19</math>. The answer is <math>1+9=10\Rightarrow \boxed{C}.</math>
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==Solution 3==
 
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n=19
 
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sum is 10 (SuperWill)
 
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==Solution 4==
 
==Solution 4==

Revision as of 17:26, 14 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

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

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

iron

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 4

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 $\boxed{n=19}$.

See Also

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. AMC logo.png

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