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

(→‎Solution 1)
m (Fixed punctuation and removed irrelevant attribution)
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\Rightarrow \ &n^2 + 4n - 437 = 0\end{split}</cmath>
 
\Rightarrow \ &n^2 + 4n - 437 = 0\end{split}</cmath>
  
−
Solving by the quadratic formula, <math>n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19</math> (since clearly <math>n \geq 0</math>). The answer is therefore <math>1 + 9 = \boxed{\textbf{(C) }10}</math>
+
Solving by the quadratic formula, <math>n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19</math> (since clearly <math>n \geq 0</math>). The answer is therefore <math>1 + 9 = \boxed{\textbf{(C) }10}</math>.
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iron
 
  
 
==Solution 2==
 
==Solution 2==
Line 22: Line 20:
 
Dividing both sides by <math>n!</math> gives
 
Dividing both sides by <math>n!</math> gives
 
<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 non-negative, <math>n=19</math>. The answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>
+
Since <math>n</math> is non-negative, <math>n=19</math>. The answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>.
  
 
==Solution 3==
 
==Solution 3==
  
−
Dividing both sides by <math>n!</math> as before gives <math>(n+1)+(n+1)(n+2)=440</math>. Now factor out <math>(n+1)</math>, giving <math>(n+1)(n+3)=440</math>. Now by considering the prime factorization of <math>440</math>, a bit of experimentation gives us <math>n+1=20</math> and <math>n+3=22</math>, so <math>n=19</math>, so the answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>
+
Dividing both sides by <math>n!</math> as before gives <math>(n+1)+(n+1)(n+2)=440</math>. Now factor out <math>(n+1)</math>, giving <math>(n+1)(n+3)=440</math>. Now by considering the prime factorization of <math>440</math>, a bit of experimentation gives us <math>n+1=20</math> and <math>n+3=22</math>, so <math>n=19</math>, so the answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>.
  
 
==Solution 4==
 
==Solution 4==
−
Obviously <math>n</math> must be very close to <math>\sqrt{440}</math>. By quick inspection, <math>n = 19</math> works, so the answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>
+
Obviously <math>n</math> must be very close to <math>\sqrt{440}</math>. By quick inspection, <math>n = 19</math> works, so the answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 19:34, 17 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

\[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\]

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 $(n+1)(n+3)=440$. Now 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

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

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

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