Difference between revisions of "2018 AMC 10A Problems/Problem 1"
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== Solution == | == Solution == | ||
− | <math>2+1=3</math> | + | We will start with <math>2+1=3</math> and then apply the operation "invert and add one" three times. These iterations yield (after <math>3</math>): <math>\frac{4}{3}</math>, <math>\frac{7}{4}</math>, and finally <math>\boxed{\textbf{(B) } \frac{11}{7} }</math> |
== See Also == | == See Also == |
Revision as of 03:11, 10 February 2018
Problem
What is the value of
Solution
We will start with and then apply the operation "invert and add one" three times. These iterations yield (after ): , , and finally
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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All AMC 10 Problems and Solutions |
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