Difference between revisions of "2018 AMC 10A Problems/Problem 1"
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== Solution == | == Solution == | ||
| + | For all nonzero numbers <math>a,</math> recall that <math>a^{-1}=\frac1a</math> is the reciprocal of <math>a.</math> | ||
| − | + | The original expression becomes | |
| + | <cmath>\begin{align*} | ||
| + | \left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 &= \left(\left(3^{-1}+1\right)^{-1}+1\right)^{-1}+1 \\ | ||
| + | &= \left(\left(\frac13+1\right)^{-1}+1\right)^{-1}+1 \\ | ||
| + | &= \left(\left(\frac43\right)^{-1}+1\right)^{-1}+1 \\ | ||
| + | &= \left(\frac34+1\right)^{-1}+1 \\ | ||
| + | &= \left(\frac74\right)^{-1}+1 \\ | ||
| + | &= \frac47+1 \\ | ||
| + | &= \boxed{\textbf{(B) }\frac{11}7}. | ||
| + | \end{align*}</cmath> | ||
| + | ~MRENTHUSIASM | ||
== Video Solutions == | == Video Solutions == | ||
Revision as of 23:49, 22 August 2021
Contents
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
For all nonzero numbers 
recall that 
is the reciprocal of 
The original expression becomes
~MRENTHUSIASM
Video Solutions
~savannahsolver
~Education, the Study of Everything
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
