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

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

2016 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 17: / Line 17: @@
 <math>\dfrac{11!-10!}{9!}</math>
 consider 10 as n
-consider the nurmetor only
+<math>\dfrac{(n+1)!-n!}{(n-1)!}</math>
-$\dfrac{11!-10!}
+simpify
-(n+1)!-n!=(n+1)n!+(-1)n!
+<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>
-= n(n!)
+subsitute n as 10 again
+<math>\dfrac{10^2}{1}</math>
-now with the deniminator
+answer is <math>10^2</math> which is 100
-n(n!)/(n-1)!
+==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
-n(n-1)!=n!
-so
-= n(n(n-1)!)/(n-1)!
-divide the (n-1)!
-we get n time n
-= n^2
-= 10^2
-= 100
 ==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==

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