2019 AMC 10B Problems/Problem 6: Difference between revisions

Revision as of 12:38, 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 $n$ ?

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

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

$\frac{-4\pm \sqrt{16+437\cdot4}}{2} \Rightarrow \frac{-4\pm 42}{2}\Rightarrow \frac{38}{2} \Rightarrow \boxed{D) 19}$

iron

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-You take the weight off of me
+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>
+==Solution==
+<math>(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!</math>
+<math>n![n+1 + (n+2)(n+1)] = 440 \cdot n!</math>
+<math>n + 1 + n^2 + 3n + 2 = 440</math>
+<math>n^2 + 4n - 437</math>
+<math>\frac{-4\pm \sqrt{16+437\cdot4}}{2} \Rightarrow \frac{-4\pm 42}{2}\Rightarrow \frac{38}{2} \Rightarrow \boxed{D) 19}</math>
+iron