Difference between revisions of "2016 AMC 10A Problems/Problem 1"
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Simplifying, we get <math>\frac{(11-1)(10!)}{9!}</math>, which equals <math>10 \cdot 10</math>. | Simplifying, we get <math>\frac{(11-1)(10!)}{9!}</math>, which equals <math>10 \cdot 10</math>. | ||
Therefore, the answer is <math>10^2</math> = <math>\boxed{\textbf{(B)}~100}</math>. | Therefore, the answer is <math>10^2</math> = <math>\boxed{\textbf{(B)}~100}</math>. | ||
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~TheGoldenRetriever | ~TheGoldenRetriever | ||
Revision as of 20:10, 10 August 2023
Contents
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
simpify
= 
= 
= 
= 
subsitute n as 10 again
answer is 
which is 100
Solution 4
This is equivalent to 
Simplifying, we get 
, which equals 
.
Therefore, the answer is = 
.
~TheGoldenRetriever
Video Solution (HOW TO THINK CREATIVELY!!!)
https://youtu.be/r5G98oPPyNM
~Education, the Study of Everything
Video Solution
~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. 
