2014 AMC 10A Problems/Problem 1
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
~savannahsolver
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
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Followed by Problem 2 | |
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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 |
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