Difference between revisions of "2018 AMC 10A Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | + | <cmath> \left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 </cmath> | |
| + | <cmath> =\left(\left(3)^{-1}+1\right)^{-1}+1\right)^{-1}+1 </cmath> | ||
| + | <cmath> =\left(\left(\frac{1}{3}+1\right)^{-1}+1\right)^{-1}+1 </cmath> | ||
| + | <cmath> =\left(\left(\frac{4}{3}\right)^{-1}+1\right)^{-1}+1 </cmath> | ||
| + | <cmath> =\left(\frac{3}{4}+1\right)^{-1}+1 </cmath> | ||
| + | <cmath> =\left(\frac{7}{4}\right)^{-1}+1 </cmath> | ||
| + | <cmath> =\frac{4}{7}+1 </cmath> | ||
| + | <cmath> =\frac{11}{7} </cmath> | ||
| + | Therefore, the answer is <math>\boxed{\textbf{(B) } \frac{11}{7} }</math>. | ||
== See Also == | == See Also == | ||
Revision as of 14:02, 13 January 2020
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
![\[\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1\]](http://latex.artofproblemsolving.com/6/b/e/6be047d20a69355e893b63ee039b7e77ca1cdba2.png)
![\[=\left(\left(3)^{-1}+1\right)^{-1}+1\right)^{-1}+1\]](http://latex.artofproblemsolving.com/b/6/6/b668f4930b1803f3772c5a39345e701ffec3c893.png)
![\[=\left(\left(\frac{1}{3}+1\right)^{-1}+1\right)^{-1}+1\]](http://latex.artofproblemsolving.com/e/5/8/e580492bb2b1c24792237da1add20a4fcf891036.png)
![\[=\left(\left(\frac{4}{3}\right)^{-1}+1\right)^{-1}+1\]](http://latex.artofproblemsolving.com/8/c/e/8cebdc51cce0b8e843fafde1aa79063f0a17c49e.png)
![\[=\left(\frac{3}{4}+1\right)^{-1}+1\]](http://latex.artofproblemsolving.com/9/6/f/96f8982c6c8641c6b807ea7193c24bd957ddeeb1.png)
![\[=\left(\frac{7}{4}\right)^{-1}+1\]](http://latex.artofproblemsolving.com/2/0/f/20fc7a854190d2a67d1af3099d1a1507e23fa0c6.png)
![\[=\frac{4}{7}+1\]](http://latex.artofproblemsolving.com/b/3/4/b34fbbdaa9c015e5b01475e570be8b3e1391fabb.png)
![\[=\frac{11}{7}\]](http://latex.artofproblemsolving.com/c/9/c/c9c2a944b021c167a14beb6ebff2b19c6dbfd571.png)
Therefore, the answer is 
.
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. 
