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

Revision as of 12:39, 14 February 2019

There is a real $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$ ?

$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$

$(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!$ $n![n+1 + (n+2)(n+1)] = 440 \cdot n!$ $n + 1 + n^2 + 3n + 2 = 440$ $n^2 + 4n - 437$

$\frac{-4\pm \sqrt{16+437\cdot4}}{2} \Rightarrow \frac{-4\pm 42}{2}\Rightarrow \frac{38}{2} \Rightarrow 1 + 9 = \boxed{C) 10}$

iron

@@ Line 4: / Line 4: @@
 ==Solution==
-<math>(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!</math>
+<cmath>(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!</cmath>
-<math>n![n+1 + (n+2)(n+1)] = 440 \cdot n!</math>
+<cmath>n![n+1 + (n+2)(n+1)] = 440 \cdot n!</cmath>
-<math>n + 1 + n^2 + 3n + 2 = 440</math>
+<cmath>n + 1 + n^2 + 3n + 2 = 440</cmath>
-<math>n^2 + 4n - 437</math>
+<cmath>n^2 + 4n - 437</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>
 iron