Difference between revisions of "2003 AMC 10A Problems/Problem 9"
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<math> \mathrm{(A) \ } \sqrt{x}\qquad \mathrm{(B) \ } \sqrt[3]{x^{2}}\qquad \mathrm{(C) \ } \sqrt[27]{x^{2}}\qquad \mathrm{(D) \ } \sqrt[54]{x}\qquad \mathrm{(E) \ } \sqrt[81]{x^{80}} </math> | <math> \mathrm{(A) \ } \sqrt{x}\qquad \mathrm{(B) \ } \sqrt[3]{x^{2}}\qquad \mathrm{(C) \ } \sqrt[27]{x^{2}}\qquad \mathrm{(D) \ } \sqrt[54]{x}\qquad \mathrm{(E) \ } \sqrt[81]{x^{80}} </math> | ||
− | + | == Solution 1 == | |
<math>\sqrt[3]{x\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{2}\cdot\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{3}}=\sqrt{x}</math>. | <math>\sqrt[3]{x\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{2}\cdot\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{3}}=\sqrt{x}</math>. | ||
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<math>\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}=\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}=\sqrt[3]{x\sqrt{x}}= \sqrt{x} \Rightarrow \boxed{\mathrm{(A)}\ \sqrt{x}}</math> | <math>\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}=\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}=\sqrt[3]{x\sqrt{x}}= \sqrt{x} \Rightarrow \boxed{\mathrm{(A)}\ \sqrt{x}}</math> | ||
− | + | == Solution 2 == | |
We know that <math>\sqrt[3]{x\sqrt{x}} = \sqrt[3]{x\cdot(x^\frac{1}{2})} = \sqrt[3]{x^\frac{3}{2}} = x^\frac{1}{2}</math> We plug this into <math>\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}</math> and get <math>\sqrt[3]{x\sqrt[3]{x\cdot(x^\frac{1}{2})}}</math>. Again, we substitute and get <math>\sqrt[3]{x\cdot(x^\frac{1}{2})}</math>. We substitute one more time and get <math>x^\frac{1}{2} = \boxed{(A) \sqrt{x}}</math>. | We know that <math>\sqrt[3]{x\sqrt{x}} = \sqrt[3]{x\cdot(x^\frac{1}{2})} = \sqrt[3]{x^\frac{3}{2}} = x^\frac{1}{2}</math> We plug this into <math>\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}</math> and get <math>\sqrt[3]{x\sqrt[3]{x\cdot(x^\frac{1}{2})}}</math>. Again, we substitute and get <math>\sqrt[3]{x\cdot(x^\frac{1}{2})}</math>. We substitute one more time and get <math>x^\frac{1}{2} = \boxed{(A) \sqrt{x}}</math>. | ||
− | + | ==Solution 3 (Lame??)== | |
− | WLOG, plug in some random square number like <math> 9 </math> or <math> 4 </math>, and the output you get after simplifying the expression will always be the square root of the number, so the answer is just \boxed{(A) \sqrt{x}} | + | WLOG, plug in some random square number like <math> 9 </math> or <math> 4 </math>, and the output you get after simplifying the expression will always be the square root of the number, so the answer is just <math> \boxed{(A) \sqrt{x}}</math>. |
+ | |||
+ | ~Darth_Cadet | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=8DMxo2pi1h8 ~David | ||
== See Also == | == See Also == |
Latest revision as of 17:47, 28 July 2023
Problem
Simplify .
Solution 1
.
Therefore:
Solution 2
We know that We plug this into and get . Again, we substitute and get . We substitute one more time and get .
Solution 3 (Lame??)
WLOG, plug in some random square number like or , and the output you get after simplifying the expression will always be the square root of the number, so the answer is just .
~Darth_Cadet
Video Solution
https://www.youtube.com/watch?v=8DMxo2pi1h8 ~David
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.