Difference between revisions of "2016 AMC 10A Problems/Problem 1"
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<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math> | <math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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> | 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> | ||
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| + | ==Solution 2== | ||
| + | We can use subtraction of fractions to get <cmath>\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!}</cmath> which will get us <math>110 -10 = \boxed{\textbf{(B)}\;100.}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 08:01, 5 February 2016
Contents
Problem
What is the value of 
?
Solution 1
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)
Solution 2
We can use subtraction of fractions to get ![\[\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!}\]](http://latex.artofproblemsolving.com/4/b/7/4b7f9aefc4bc405d52960b5bf8683575f9b88127.png)
which will get us 
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. 
