Difference between revisions of "2016 AMC 10A Problems/Problem 1"
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This is equivalent to <math>\frac{11(10!) - 1(10!)}{9!}</math> | This is equivalent to <math>\frac{11(10!) - 1(10!)}{9!}</math> | ||
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>. | ||
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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>. | ||
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
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 = .
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Video Solution
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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.