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 13:30, 28 September 2019
Problem
What is the value of
![\[\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?\]](http://latex.artofproblemsolving.com/5/8/f/58f83d4d7d692d9e5ae0dbd561e9e28ef0808623.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
