Difference between revisions of "2018 AMC 10A Problems/Problem 1"
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<cmath>\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?</cmath><math>\textbf{(A) } \frac58 \qquad \textbf{(B) }\frac{11}7 \qquad \textbf{(C) } \frac85 \qquad \textbf{(D) } \frac{18}{11} \qquad \textbf{(E) } \frac{15}8 </math> | <cmath>\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?</cmath><math>\textbf{(A) } \frac58 \qquad \textbf{(B) }\frac{11}7 \qquad \textbf{(C) } \frac85 \qquad \textbf{(D) } \frac{18}{11} \qquad \textbf{(E) } \frac{15}8 </math> | ||
| − | == Solution | + | == Solution == |
Arithmetic gives <math>(\textbf{B})</math> | Arithmetic gives <math>(\textbf{B})</math> | ||
== See Also == | == See Also == | ||
| − | {{AMC10 box|year=2018|ab=A | + | {{AMC10 box|year=2018|ab=A|num-a=2}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:34, 8 February 2018
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
Arithmetic gives 
See Also
| 2018 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by [[2018 AMC 10A Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]] |
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. 
