Difference between revisions of "2016 AMC 10A Problems/Problem 1"
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<math>\boxed{\textbf{(B)}~100}</math> | <math>\boxed{\textbf{(B)}~100}</math> | ||
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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> | 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 21:12, 27 September 2022
Problem
What is the value of ?
Solution 1
Solution 1
We can use subtraction of fractions to get
Solution 3
Factoring out gives .
Video Solution
~IceMatrix
~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.