2018 AMC 10A Problems/Problem 1: Difference between revisions
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== Problem == | |||
What is the value of | What is the value of | ||
<cmath>\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?</cmath | <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 == | |||
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 Solution (Simplicity)== | |||
https://www.youtube.com/watch?v=g0o4BbECwlk&t=16s | |||
==Video Solution (HOW TO THINK CREATIVELY!)== | |||
https://youtu.be/19mpsCcQzY0 | |||
~Education, the Study of Everything | |||
== Video Solutions == | |||
https://youtu.be/vO-ELYmgRI8 | |||
https://youtu.be/cat3yTIpX4k | |||
~savannahsolver | |||
== See Also == | == See Also == | ||
{{AMC10 box|year=2018|ab=A|before=First Problem|num-a=2}} | {{AMC10 box|year=2018|ab=A|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:46, 26 December 2023
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 Solution (Simplicity)
https://www.youtube.com/watch?v=g0o4BbECwlk&t=16s
Video Solution (HOW TO THINK CREATIVELY!)
~Education, the Study of Everything
Video Solutions
~savannahsolver
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. Error creating thumbnail: File missing
