Difference between revisions of "2018 AMC 10A Problems/Problem 1"
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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> |
− | == Solution == | + | <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| | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:46, 26 December 2023
Contents
Problem
What is the value of
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.