Difference between revisions of "2016 AMC 10A Problems/Problem 1"
(→Solution) |
(→Solution) |
||
| Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
| − | Factoring out the <math>10!</math> from the numerator and cancelling out the <math>9!</math>s in the numerator and denominator, we have: <cmath>\frac{11!-10!}{9!} = \frac{9!( | + | Factoring out the <math>10!</math> from the numerator and cancelling out the <math>9!</math>s in the numerator and denominator, we have: <cmath>\frac{11!-10!}{9!} = \frac{11 \cdot 10! - 1 \cdot 10!}{9!} = \frac{(11 - 1) \cdot (10!)}{9!} = 10 \cdot 10 =\boxed{\textbf{(B)}\;100.}</cmath> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}} | {{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:15, 3 February 2016
Problem
What is the value of 
?
Solution
Factoring out the 
from the numerator and cancelling out the 
s in the numerator and denominator, we have: ![\[\frac{11!-10!}{9!} = \frac{11 \cdot 10! - 1 \cdot 10!}{9!} = \frac{(11 - 1) \cdot (10!)}{9!} = 10 \cdot 10 =\boxed{\textbf{(B)}\;100.}\]](http://latex.artofproblemsolving.com/3/f/4/3f442bb96fda08b72dcb27a2cc662f01c9be452a.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
