Difference between revisions of "2016 AMC 10A Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
| − | Factoring out | + | Factoring out <math>10!</math> from the numerator and cancelling out <math>9!</math> from the numerator and the denominator, we have <cmath>\frac{11!-10!}{9!} = \frac{11 \cdot 10! - 1 \cdot 10!}{9!} = \frac{(10!) \cdot (11 - 1)}{9!} = 10 \cdot 10 =\boxed{\textbf{(B)}\;100}.</cmath> |
==Solution 2== | ==Solution 2== | ||
| − | We can use subtraction of fractions to get <cmath>\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} | + | |
| + | We can use subtraction of fractions to get <cmath>\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{\textbf{(B)}\;100}.</cmath> | ||
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==Solution 3== | ==Solution 3== | ||
Revision as of 17:36, 17 February 2016
Problem
What is the value of 
?
Solution 1
Factoring out 
from the numerator and cancelling out 
from the numerator and the denominator, we have ![\[\frac{11!-10!}{9!} = \frac{11 \cdot 10! - 1 \cdot 10!}{9!} = \frac{(10!) \cdot (11 - 1)}{9!} = 10 \cdot 10 =\boxed{\textbf{(B)}\;100}.\]](http://latex.artofproblemsolving.com/9/6/8/968bf125a0dbe448399937e8d1dab7b8c2e88643.png)
Solution 2
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 3
Factoring out gives 
.
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. 
