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

(Solution 1)
(Video Solution)
 
(18 intermediate revisions by 11 users not shown)
Line 3: Line 3:
 
==Problem==
 
==Problem==
  
There is a real <math>n</math> such that <math>(n+1)! + (n+2)! = n! \cdot 440</math>. What is the sum of the digits of <math>n</math>?
+
There is a positive integer <math>n</math> such that <math>(n+1)! + (n+2)! = n! \cdot 440</math>. What is the sum of the digits of <math>n</math>?
  
 
<math>\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12</math>
 
<math>\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12</math>
  
==Solution 1==
+
==Solution==
 +
===Solution 1===
  
<cmath>(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!</cmath>
+
<cmath>\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\
<cmath>n![n+1 + (n+2)(n+1)] = 440 \cdot n!</cmath>
+
\Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\
<cmath>n + 1 + n^2 + 3n + 2 = 440</cmath>
+
\Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\
<cmath>n^2 + 4n - 437 = 0</cmath>
+
\Rightarrow \ &n^2 + 4n - 437 = 0\end{split}</cmath>
  
<math>\frac{-4\pm \sqrt{16+437\cdot4}}{2} \Rightarrow \frac{-4\pm 42}{2}\Rightarrow \frac{38}{2} \Rightarrow 19</math>. <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>.
  
==Solution 2==
+
===Solution 2===
  
 
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 positive, <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===
  
Divide both sides by <math>n!</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>. 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>.
  
 +
== Video Solution ==
 +
https://youtu.be/ba6w1OhXqOQ?t=1956
  
<math>(n+1)+(n+1)(n+2)=440</math>
+
~ pi_is_3.14
  
factor out <math>(n+1)</math>:
+
==Video Solution==
 +
https://youtu.be/7xf_g3YQk00
  
<math>(n+1)*(n+3)=440</math>
+
~IceMatrix
  
 +
https://youtu.be/6YFN_hwotUk
  
prime factorization of <math>440</math> and 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>
+
~savannahsolver
 
 
==Solution 4==
 
Obviously n must be very close to <math>\sqrt{440}</math>. By quick inspection, <math>n = 19</math> works. <math>1 + 9 = \boxed{\textbf{(C) }10}</math>
 
  
 
==See Also==
 
==See Also==

Latest revision as of 03:26, 15 January 2021

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

Video Solution

https://youtu.be/ba6w1OhXqOQ?t=1956

~ pi_is_3.14

Video Solution

https://youtu.be/7xf_g3YQk00

~IceMatrix

https://youtu.be/6YFN_hwotUk

~savannahsolver

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

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

Invalid username
Login to AoPS