2016 AMC 10A Problems/Problem 1

Revision as of 13:41, 17 March 2023 by Imaginary1234 (talk | contribs) (Solution 3)

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 consider the nurmetor only $\dfrac{11!-10!} (n+1)!-n!=(n+1)n!+(-1)n! = n(n!)

now with the deniminator

n(n!)/(n-1)!

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

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

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