2013 AMC 12A Problems/Problem 4
Contents
[hide]Problem
What is the value of
Solution
We can factor a out of the numerator and denominator to obtain
The cancels, so we get
, which is
Solution 2
Suppose that Then the given expression is equal to
~sugar_rush
Solution 3 (Elimination)
Choice cannot be true, because is clearly larger than . We can apply our same logic to choice and eliminate it as well. and are all whole numbers, but is not a multiple of , so we can eliminate choices and too. This leaves us with choice as our final answer.
~dbnl
Video Solution
https://www.youtube.com/watch?v=2vf843cvVzo?t=545 (solution for #4 starts at 9:05)
~sugar_rush
See also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 |
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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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