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

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<math>\frac{11!-10!}{9!}=\frac{11\cdot10!-10!}{9!}=\frac{100\cdot9!}{9!}=100</math>
 
<math>\frac{11!-10!}{9!}=\frac{11\cdot10!-10!}{9!}=\frac{100\cdot9!}{9!}=100</math>
<math>B</math>
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<math>\boxed{\textbf{(B)}~100}</math>
  
 
==Solution 2==
 
==Solution 2==
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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>.
 
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>.
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==Video Solution==
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https://youtu.be/VIt6LnkV4_w
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~IceMatrix
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https://youtu.be/CrS7oHDrvP8
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 17:16, 16 June 2020

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

$\frac{11!-10!}{9!}=\frac{11\cdot10!-10!}{9!}=\frac{100\cdot9!}{9!}=100$


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

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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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