Difference between revisions of "2017 AMC 8 Problems/Problem 3"
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− | ==Problem | + | ==Problem== |
What is the value of the expression <math>\sqrt{16\sqrt{8\sqrt{4}}}</math>? | What is the value of the expression <math>\sqrt{16\sqrt{8\sqrt{4}}}</math>? | ||
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<math>\textbf{(A) }4\qquad\textbf{(B) }4\sqrt{2}\qquad\textbf{(C) }8\qquad\textbf{(D) }8\sqrt{2}\qquad\textbf{(E) }16</math> | <math>\textbf{(A) }4\qquad\textbf{(B) }4\sqrt{2}\qquad\textbf{(C) }8\qquad\textbf{(D) }8\sqrt{2}\qquad\textbf{(E) }16</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | + | <math>\sqrt{16\sqrt{8\sqrt{4}}}</math> = <math>\sqrt{16\sqrt{8\cdot 2}}</math> = <math>\sqrt{16\sqrt{16}}</math> = <math>\sqrt{16\cdot 4}</math> = <math>\sqrt{64}</math> = <math>\boxed{\textbf{(C)}\ 8}</math>. | |
− | ~ | + | ==Worse Solution== |
+ | ~ Sahan | ||
+ | |||
+ | We solve the general form expression <math>\sqrt{a\sqrt{b\sqrt{c}}}</math>. Note, | ||
+ | <cmath>\sqrt{a\sqrt{b\sqrt{c}}}=(a^4b^2c^1)^\frac{1}{8}</cmath> | ||
+ | Thus our answer is, | ||
+ | <cmath>(16^4\cdot8^24^1)^\frac{1}{8}=16777216^{\frac{1}{8}}=8</cmath> | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/elN5lYfeKnw | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/cY4NYSAD0vQ | ||
+ | |||
+ | https://youtu.be/H0WHiLy1cFg | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 19:13, 4 July 2024
Contents
Problem
What is the value of the expression ?
Solution 1
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=
=
=
.
Worse Solution
~ Sahan
We solve the general form expression . Note,
Thus our answer is,
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.