Difference between revisions of "2016 AMC 10A Problems/Problem 1"

Revision as of 13:40, 17 March 2023

What is the value of $\dfrac{11!-10!}{9!}$ ?

$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$

We can use subtraction of fractions to get $\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{\textbf{(B)}\;100}.$

Factoring out $9!$ gives $\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{\textbf{(B)}~100}$ .

consider 10 as n consider the nurmetor only (n+1)!-n!=(n+1)n!+(-1)n! = n(n!)

now with the deniminator

n(n!)/(n-1)!

n(n-1)!=n!

so = n(n(n-1)!)/(n-1)! divide the (n-1)! we get n time n = n^2 = 10^2 = 100

~IceMatrix

~savannahsolver

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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

@@ Line 12: / Line 12: @@
 Factoring out <math>9!</math> gives <math>\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{\textbf{(B)}~100}</math>.
+==Solution 3==
+consider 10 as n
+consider the nurmetor only
+(n+1)!-n!=(n+1)n!+(-1)n!
+= n(n!)
+now with the deniminator
+n(n!)/(n-1)!
+n(n-1)!=n!
+so
+= n(n(n-1)!)/(n-1)!
+divide the (n-1)!
+we get n time n
+= n^2
+= 10^2
+= 100
 ==Video Solution==