2019 AMC 10B Problems/Problem 6
- 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
![\[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\]](http://latex.artofproblemsolving.com/e/9/4/e94a76dd637eee784abfc844bcc187bc69d72464.png)
Solving by the quadratic formula, 
(since clearly 
). The answer is therefore 
.
~IronicNinja
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 
. Now 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 | |
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