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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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Factoring out <math>10!</math> from the numerator and cancelling out <math>9!</math> from the numerator and the denominator, we have <cmath>\frac{11!-10!}{9!} = \frac{11 \cdot 10! - 1 \cdot 10!}{9!} =  \frac{(10!) \cdot (11 - 1)}{9!} = 10 \cdot 10 =\boxed{\textbf{(B)}\;100}.</cmath>
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We can use subtraction of fractions to get <cmath>\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{\textbf{(B)}\;100}.</cmath>
  
 
==Solution 2==
 
==Solution 2==
  
−
We can use subtraction of fractions to get <cmath>\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{\textbf{(B)}\;100}.</cmath>
+
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==
 
==Solution 3==
 +
<math>\dfrac{11!-10!}{9!}</math>
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consider 10 as n
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<math>\dfrac{(n+1)!-n!}{(n-1)!}</math>
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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>
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 +
answer is <math>10^2</math> which is 100
 +
 +
==Solution 4==
 +
We are given the equation <math>\frac{11!-10!}{9!}</math>
 +
 +
This is equivalent to <math>\frac{11(10!) - 1(10!)}{9!}</math>
 +
Simplifying, we get <math>\frac{(11-1)(10!)}{9!}</math>, which equals <math>10 \cdot 10</math>.
 +
 +
Therefore, the answer is <math>10^2</math> = <math>\boxed{\textbf{(B)}~100}</math>.
 +
 +
~TheGoldenRetriever
 +
 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/r5G98oPPyNM
 +
 +
~Education, the Study of Everything
 +
 +
 +
 +
==Video Solution==
 +
https://youtu.be/VIt6LnkV4_w
 +
 +
 +
https://youtu.be/CrS7oHDrvP8
 +
 +
~savannahsolver
 +
 +
==Video Solution (FASTEST METHOD!)==
 +
 +
https://youtu.be/jowREGsZaTs
 +
 +
~Veer Mahajan
  
−
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>.
 
  
 
==See Also==
 
==See Also==
−
{{AMC10 box|year=2016|ab=A|before=|num-a=2}}
+
{{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}}
−
{{AMC12 box|year=2016|ab=A|before=|num-a=2}}
+
{{AMC12 box|year=2016|ab=A|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:59, 1 November 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

Solution 4

We are given the equation $\frac{11!-10!}{9!}$

This is equivalent to $\frac{11(10!) - 1(10!)}{9!}$ Simplifying, we get $\frac{(11-1)(10!)}{9!}$, which equals $10 \cdot 10$.

Therefore, the answer is $10^2$ = $\boxed{\textbf{(B)}~100}$.

~TheGoldenRetriever

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/r5G98oPPyNM

~Education, the Study of Everything


Video Solution

https://youtu.be/VIt6LnkV4_w


https://youtu.be/CrS7oHDrvP8

~savannahsolver

Video Solution (FASTEST METHOD!)

https://youtu.be/jowREGsZaTs

~Veer Mahajan


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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