2020 AMC 12B Problems/Problem 6
Contents
Problem
For all integers the value of is always which of the following?
Solution 1
We first expand the expression: We can now divide out a common factor of from each term of the numerator: Factoring out we get which proves that the answer is
Solution 2
In the numerator, we factor out an to get Now, without loss of generality, test values of until only one answer choice is left valid:
- knocking out and
- knocking out
This leaves as the only answer choice left.
This solution does not consider the condition The reason is that, with further testing it becomes clear that for all we get as proved in Solution 1. The condition was added most likely to encourage picking and discourage substituting smaller values into
~DBlack2021 (Solution)
~MRENTHUSIASM (Edits in Logic)
~Countmath1 (Minor Edits in Formatting)
Video Solution by Omega Learn
https://youtu.be/ba6w1OhXqOQ?t=2234
~ pi_is_3.14
Video Solution
~IceMatrix
Video Solution
~Education, the Study of Everything
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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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