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Difference between revisions of "2016 AMC 10A Problems/Problem 1"
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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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==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
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<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>
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subsitute n as 10 again 
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<math>\dfrac{10^2}{1}</math>
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answer is <math>10^2</math> which is <math>\boxed{\textbf{(B)}~100}</math>.
 +
 +
==Solution 4==
 +
We are given the equation <math>\frac{11!-10!}{9!}</math>
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 +
This is equivalent to <math>\frac{11(10!) - 1(10!)}{9!}</math>
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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>.
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 +
~TheGoldenRetriever
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 +
==Solution 5 (This is a joke)==
 +
 +
Let
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<cmath>
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I_n := \int_0^\infty x^n e^{-x} \, dx = n!
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</cmath>
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(the Gamma–integral).
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 +
Then
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<cmath>
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\frac{11! - 10!}{9!} = \frac{I_{11} - I_{10}}{I_9} = \frac{\int_0^\infty e^{-x} (x^{11} - x^{10}) \, dx}{I_9}.
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</cmath>
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Integration by parts on the numerator with \( u = x^{11} - x^{10} \), \( dv = e^{-x} \, dx \) (so \( du = (11x^{10} - 10x^9) \, dx \), \( v = -e^{-x} \)) gives
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<cmath>
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\int_0^\infty e^{-x} (x^{11} - x^{10}) \, dx = \left[ -e^{-x} (x^{11} - x^{10}) \right]_0^\infty + \int_0^\infty e^{-x} (11x^{10} - 10x^9) \, dx = 11 I_{10} - 10 I_9,
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</cmath>
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since the boundary term vanishes.
 +
 +
Hence
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<cmath>
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\frac{I_{11} - I_{10}}{I_9} = \frac{11 I_{10}}{I_9} - 10.
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</cmath>
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Do another integration by parts to relate \( I_{10} \) and \( I_9 \):
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<cmath>
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I_{10} = \int_0^\infty x^{10} e^{-x} \, dx = \left[ -x^{10} e^{-x} \right]_0^\infty + 10 \int_0^\infty x^9 e^{-x} \, dx = 10 I_9.
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</cmath>
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Therefore
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<cmath>
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\frac{I_{11} - I_{10}}{I_9} = 11 \cdot 10 - 10 = 100.
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</cmath>
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<math>\boxed{(B) 100}</math>.
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 +
~Pinotation
 +
 +
~Minorly Edited by OffBrandCab
 +
 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/r5G98oPPyNM
 +
 +
~Education, the Study of Everything
 +
 +
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/VIt6LnkV4_w
 
https://youtu.be/VIt6LnkV4_w
  
−
~IceMatrix
 
  
 
https://youtu.be/CrS7oHDrvP8
 
https://youtu.be/CrS7oHDrvP8
  
 
~savannahsolver
 
~savannahsolver
 +
 +
==Video Solution (FASTEST METHOD!)==
 +
 +
https://youtu.be/jowREGsZaTs
 +
 +
~Veer Mahajan
 +
  
 
==See Also==
 
==See Also==
Line 26: Line 98:
 
{{AMC12 box|year=2016|ab=A|before=First Problem|num-a=2}}
 
{{AMC12 box|year=2016|ab=A|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category: Introductory Algebra Problems]]

Latest revision as of 22:22, 26 October 2025

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

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

Solution 5 (This is a joke)

Let \[I_n := \int_0^\infty x^n e^{-x} \, dx = n!\] (the Gamma–integral).

Then \[\frac{11! - 10!}{9!} = \frac{I_{11} - I_{10}}{I_9} = \frac{\int_0^\infty e^{-x} (x^{11} - x^{10}) \, dx}{I_9}.\] Integration by parts on the numerator with \( u = x^{11} - x^{10} \), \( dv = e^{-x} \, dx \) (so \( du = (11x^{10} - 10x^9) \, dx \), \( v = -e^{-x} \)) gives \[\int_0^\infty e^{-x} (x^{11} - x^{10}) \, dx = \left[ -e^{-x} (x^{11} - x^{10}) \right]_0^\infty + \int_0^\infty e^{-x} (11x^{10} - 10x^9) \, dx = 11 I_{10} - 10 I_9,\] since the boundary term vanishes.

Hence \[\frac{I_{11} - I_{10}}{I_9} = \frac{11 I_{10}}{I_9} - 10.\] Do another integration by parts to relate \( I_{10} \) and \( I_9 \): \[I_{10} = \int_0^\infty x^{10} e^{-x} \, dx = \left[ -x^{10} e^{-x} \right]_0^\infty + 10 \int_0^\infty x^9 e^{-x} \, dx = 10 I_9.\] Therefore \[\frac{I_{11} - I_{10}}{I_9} = 11 \cdot 10 - 10 = 100.\]

$\boxed{(B) 100}$.

~Pinotation

~Minorly Edited by OffBrandCab

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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