Difference between revisions of "2018 AMC 10A Problems/Problem 1"
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== Solution == | == Solution == | ||
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> | 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> | ||
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== See Also == | == See Also == |
Revision as of 14:30, 28 September 2019
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 | |
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 |
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