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

(Solution 4)
m (Solution 4)
Line 29: Line 29:
 
This is equivalent to <math>\frac{11(10!) - 1(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>.
 
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>.
 
Therefore, the answer is <math>10^2</math> = <math>\boxed{\textbf{(B)}~100}</math>.
  

Revision as of 20:10, 10 August 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

$\frac{11!-10!}{9!}$ 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

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

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