Difference between revisions of "2000 AIME II Problems/Problem 1"
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&= {\frac{1}{6}}.\end{align*}</cmath> | &= {\frac{1}{6}}.\end{align*}</cmath> | ||
− | Therefore our answer is <math>1 + 6 = \boxed{ | + | Therefore our answer is <math>1 + 6 = \boxed{007}</math>. |
{{AIME box|year=2000|n=II|before=First Question|num-a=2}} | {{AIME box|year=2000|n=II|before=First Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:57, 4 December 2022
Contents
[hide]Problem
The number

can be written as where
and
are relatively prime positive integers. Find
.
Solution
Solution 1
Therefore,
Solution 2
Alternatively, we could've noted that, because
Therefore our answer is .
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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All AIME Problems and Solutions |
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