Difference between revisions of "2016 AMC 10A Problems/Problem 1"
m (→Solution 4) |
Veerkmahajan (talk | contribs) |
||
(One intermediate revision by one other user not shown) | |||
Line 26: | Line 26: | ||
==Solution 4== | ==Solution 4== | ||
− | <math>\frac{11!-10!}{9!}</math> | + | We are given the equation <math>\frac{11!-10!}{9!}</math> |
+ | |||
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>. | ||
Line 48: | Line 49: | ||
~savannahsolver | ~savannahsolver | ||
+ | |||
+ | ==Video Solution (FASTEST METHOD!)== | ||
+ | |||
+ | https://youtu.be/jowREGsZaTs | ||
+ | |||
+ | ~Veer Mahajan | ||
+ | |||
==See Also== | ==See Also== |
Latest revision as of 21:59, 1 November 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
We are given the equation
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
Video Solution (FASTEST METHOD!)
~Veer Mahajan
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.