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

Latest revision as of 01:08, 28 August 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 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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 7: / Line 7: @@
 ==Solution 1==
-<math>\frac{1}{2}=\frac{1}{2}=\frac{1}{2}=\frac{1}{2}</math>
+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==
+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>.
-<math>\boxed{\textbf{(B)}~100}</math>
-==Solution 2==
+==Solution 3==
+<math>\dfrac{11!-10!}{9!}</math>
+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 <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
+==Solution 5 (This is a joke)==
+Let
+<cmath>
+I_n := \int_0^\infty x^n e^{-x} \, dx = n!
+</cmath>
+(the Gamma–integral).
+Then
+<cmath>
+\frac{11! - 10!}{9!} = \frac{I_{11} - I_{10}}{I_9} = \frac{\int_0^\infty e^{-x} (x^{11} - x^{10}) \, dx}{I_9}.
+</cmath>
+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
+<cmath>
+\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,
+</cmath>
+since the boundary term vanishes.
+Hence
+<cmath>
+\frac{I_{11} - I_{10}}{I_9} = \frac{11 I_{10}}{I_9} - 10.
+</cmath>
+Do another integration by parts to relate \( I_{10} \) and \( I_9 \):
+<cmath>
+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.
+</cmath>
+Therefore
+<cmath>
+\frac{I_{11} - I_{10}}{I_9} = 11 \cdot 10 - 10 = 100.
+</cmath>
+<math>\boxed{(B) 100}</math>.
+~Pinotation
+~Minorly Edited by OffBrandCab
-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>
+==Video Solution (HOW TO THINK CREATIVELY!!!)==
+ https://youtu.be/r5G98oPPyNM
+~Education, the Study of Everything
-==Solution 3==
-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>.
 ==Video Solution==
 https://youtu.be/VIt6LnkV4_w
-~IceMatrix
 https://youtu.be/CrS7oHDrvP8
 ~savannahsolver
+==Video Solution (FASTEST METHOD!)==
+https://youtu.be/jowREGsZaTs
+~Veer Mahajan
 ==See Also==

Page

Toolbox

Search