Difference between revisions of "2014 AMC 10A Problems/Problem 1"
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==Video Solution== | ==Video Solution== |
Latest revision as of 23:12, 26 June 2023
Contents
Problem
What is
Solution
We have Making the denominators equal gives Finally, simplifying gives
Solution 2
We have By Distributive Property, Now, we want to find the least common multiple of and so Converting everything to a denominator of Now, we use Euclidean Algorithm, to find if this fraction is reducible, so Thus, both the numerator and denominator are divisible by so
- kante314
Video Solution (CREATIVE THINKING)
https://youtu.be/sbz01QUWY6A
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
2014 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 |
2014 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.