Difference between revisions of "2019 AMC 10B Problems/Problem 6"
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<cmath>n![n+1 + (n+2)(n+1)] = 440 \cdot n!</cmath> | <cmath>n![n+1 + (n+2)(n+1)] = 440 \cdot n!</cmath> | ||
<cmath>n + 1 + n^2 + 3n + 2 = 440</cmath> | <cmath>n + 1 + n^2 + 3n + 2 = 440</cmath> | ||
| − | <cmath>n^2 + 4n - 437</cmath> | + | <cmath>n^2 + 4n - 437 = 0</cmath> |
<math>\frac{-4\pm \sqrt{16+437\cdot4}}{2} \Rightarrow \frac{-4\pm 42}{2}\Rightarrow \frac{38}{2} \Rightarrow 1 + 9 = \boxed{C) 10}</math> | <math>\frac{-4\pm \sqrt{16+437\cdot4}}{2} \Rightarrow \frac{-4\pm 42}{2}\Rightarrow \frac{38}{2} \Rightarrow 1 + 9 = \boxed{C) 10}</math> | ||
iron | iron | ||
Revision as of 13:43, 14 February 2019
There is a real 
such that 
. What is the sum of the digits of ?
Solution
![\[(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!\]](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
