Difference between revisions of "2016 AMC 10A Problems/Problem 1"
(→Solution 3) |
(→Solution 3) |
||
| Line 18: | Line 18: | ||
consider 10 as n | consider 10 as n | ||
consider the nurmetor only | consider the nurmetor only | ||
| + | $\dfrac{11!-10!} | ||
(n+1)!-n!=(n+1)n!+(-1)n! | (n+1)!-n!=(n+1)n!+(-1)n! | ||
= n(n!) | = n(n!) | ||
Revision as of 14:41, 17 March 2023
Problem
What is the value of 
?
Solution 1
We can use subtraction of fractions to get ![\[\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{\textbf{(B)}\;100}.\]](http://latex.artofproblemsolving.com/5/d/8/5d8e4e6c76063dc4767a679ab11d828b652b9470.png)
Solution 2
Factoring out 
gives 
.
Solution 3
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
~IceMatrix
~savannahsolver
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
