Difference between revisions of "2018 AMC 10A Problems/Problem 1"

Latest revision as of 20:46, 26 December 2023

Problem

What is the value of $\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?$

$\textbf{(A) } \frac58 \qquad \textbf{(B) }\frac{11}7 \qquad \textbf{(C) } \frac85 \qquad \textbf{(D) } \frac{18}{11} \qquad \textbf{(E) } \frac{15}8$

Solution

For all nonzero numbers $a,$ recall that $a^{-1}=\frac1a$ is the reciprocal of $a.$

The original expression becomes $\begin{align*} \left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 &= \left(\left(3^{-1}+1\right)^{-1}+1\right)^{-1}+1 \\ &= \left(\left(\frac13+1\right)^{-1}+1\right)^{-1}+1 \\ &= \left(\left(\frac43\right)^{-1}+1\right)^{-1}+1 \\ &= \left(\frac34+1\right)^{-1}+1 \\ &= \left(\frac74\right)^{-1}+1 \\ &= \frac47+1 \\ &= \boxed{\textbf{(B) }\frac{11}7}. \end{align*}$ ~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

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.

@@ Line 1: / Line 1: @@
+== Problem ==
 What is the value of
-<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 ==
+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
-===Solution===
-Evaluating the expressions starting with the innermost one, we get that the answer is <math>\boxed{\textbf{(B)}\ \frac{11}{7}}</math>
 == See Also ==
+{{AMC10 box|year=2018|ab=A|before=First Problem|num-a=2}}
-{{AMC10 box|year=2017|ab=A|before=First Problem|num-a=2}}
 {{MAA Notice}}

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