Difference between revisions of "2018 AMC 10A Problems/Problem 1"
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== Solution == | == Solution == | ||
<math>2+1=3</math>, so the reciprocal is <math>\frac{1}{3}</math>. Summing <math>1</math> and repeating leads to <math>\frac{3}{4}</math>. Rinse and repeat to get <math>\frac{4}{7}</math>. Adding <math>1</math> to that is <math>\frac{11}{7}</math> (B) | <math>2+1=3</math>, so the reciprocal is <math>\frac{1}{3}</math>. Summing <math>1</math> and repeating leads to <math>\frac{3}{4}</math>. Rinse and repeat to get <math>\frac{4}{7}</math>. Adding <math>1</math> to that is <math>\frac{11}{7}</math> (B) | ||
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== See Also == | == See Also == | ||
Revision as of 21:57, 8 February 2018
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
, so the reciprocal is 
. Summing 
and repeating leads to 
. Rinse and repeat to get 
. Adding to that is 
(B)
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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