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

Revision as of 13:45, 17 March 2023

Problem

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

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

Solution 1

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

Solution 2

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

Solution 3

$\dfrac{11!-10!}{9!}$ consider 10 as n $\dfrac{(n+1)!-n!}{(n-1)!}$ simpify $\dfrac{(n+1)n!-(-1)n!}{(n-1)!}$ = $\dfrac{n(n!)}{(n-1)!}$ = $\dfrac{n(n(n-1)!)}{(n-1)!}$ = $\dfrac{n(n)(1)}{(1}$ = $\dfrac{n^2}{1}$ subsitute n as 10 again $\dfrac{10^2}{1}$

answer is 10^2 which is 100

Video Solution

https://youtu.be/VIt6LnkV4_w

~IceMatrix

https://youtu.be/CrS7oHDrvP8

~savannahsolver

2016 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2016 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 18: / Line 18: @@
 consider 10 as n
 <math>\dfrac{(n+1)!-n!}{(n-1)!}</math>
+simpify
+<math>\dfrac{(n+1)n!-(-1)n!}{(n-1)!}</math> = <math>\dfrac{n(n!)}{(n-1)!}</math> = <math>\dfrac{n(n(n-1)!)}{(n-1)!}</math> = <math>\dfrac{n(n)(1)}{(1}</math> = <math>\dfrac{n^2}{1}</math>
+subsitute n as 10 again
+<math>\dfrac{10^2}{1}</math>
+answer is 10^2 which is 100
 ==Video Solution==

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