Difference between revisions of "2019 AMC 10B Problems/Problem 6"
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| + | {{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #6]] and [[2019 AMC 12B Problems|2019 AMC 12B #4]]}} | ||
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| + | ==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 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>? | ||
<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== | + | ==Solution 1== |
| + | |||
<cmath>(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!</cmath> | <cmath>(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!</cmath> | ||
<cmath>n![n+1 + (n+2)(n+1)] = 440 \cdot n!</cmath> | <cmath>n![n+1 + (n+2)(n+1)] = 440 \cdot n!</cmath> | ||
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iron | iron | ||
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| + | ==Solution 2== | ||
| + | |||
| + | 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> | ||
| + | Since <math>n</math> is positive, <math>n=19</math>. The answer is <math>1+9=10\Rightarrow \boxed{C}.</math> | ||
| + | |||
| + | ==Solution 3== | ||
| + | n=19 | ||
| + | sum is 10 (SuperWill) | ||
| + | |||
| + | |||
| + | ==Solution 4== | ||
| + | |||
| + | Divide both sides by <math>n!</math>: | ||
| + | |||
| + | |||
| + | <math>(n+1)+(n+1)(n+2)=440</math> | ||
| + | |||
| + | factor out <math>(n+1)</math>: | ||
| + | |||
| + | <math>(n+1)*(n+3)=440</math> | ||
| + | |||
| + | |||
| + | 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>\boxed{n=19}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2019|ab=B|num-b=5|num-a=7}} | ||
| + | {{AMC12 box|year=2019|ab=B|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Revision as of 14:28, 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 
such that 
. What is the sum of the digits of ?
Solution 1
![\[(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!\]](http://latex.artofproblemsolving.com/8/8/8/8881090c2d4ba110904d1101dc062812232ec0d5.png)
![\[n![n+1 + (n+2)(n+1)] = 440 \cdot n!\]](http://latex.artofproblemsolving.com/7/b/a/7ba39b77bb6dd2b4e6936d960ad4c3d4f0cf9a1a.png)
![\[n + 1 + n^2 + 3n + 2 = 440\]](http://latex.artofproblemsolving.com/7/2/9/7295d688733637ef2be66462eb5589580892ff54.png)
![\[n^2 + 4n - 437 = 0\]](http://latex.artofproblemsolving.com/3/9/6/3965747c92a22e615bff8ab192c20dee20372321.png)
. 
iron
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 positive, 
. The answer is 
Solution 3
n=19 sum is 10 (SuperWill)
Solution 4
Divide both sides by :
factor out 
:
prime factorization of 
and a bit of experimentation gives us 
and 
, so 
.
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. 
