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

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==Solution 1==
 
==Solution 1==
  
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<math>\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2} =\frac{1}{2}=\frac{1}{2}=\frac{1}{2}\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2} </math>
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<math>\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2} =\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2} </math>
  
  

Revision as of 19:44, 26 September 2022

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

$\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2} =\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}$


$\boxed{\textbf{(B)}~100}$

Solution 2

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 3

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

Video Solution

https://youtu.be/VIt6LnkV4_w

~IceMatrix

https://youtu.be/CrS7oHDrvP8

~savannahsolver

See Also

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

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