Difference between revisions of "2019 AMC 10B Problems/Problem 6"
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<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>\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ | <cmath>\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ | ||
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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>. | ||
| − | ==Solution 2== | + | ===Solution 2=== |
Dividing both sides by <math>n!</math> gives | Dividing both sides by <math>n!</math> gives | ||
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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>. 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>. 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>. | ||
Revision as of 18:14, 1 February 2020
- 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 
such that 
. What is the sum of the digits of ?
Solution
Solution 1
![\[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\]](http://latex.artofproblemsolving.com/e/9/4/e94a76dd637eee784abfc844bcc187bc69d72464.png)
Solving by the quadratic formula, 
(since clearly 
). The answer is therefore 
.
Solution 2
Dividing both sides by 
gives
![\[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.\]](http://latex.artofproblemsolving.com/9/5/a/95acebc265a9f6c6779a1c3e16f867517e875b6a.png)
Since is non-negative, 
. The answer is .
Solution 3
Dividing both sides by as before gives 
. Now factor out 
, giving 
. By considering the prime factorization of 
, a bit of experimentation gives us 
and 
, so , so the answer is .
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. 
