Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1956 AHSME Problems/Problem 9"

(Created page with "== Solution == Notice that the first part and the second part of the expression are the same things, just in a different order, ie 3 and the 6 are switched around. Therefore,...")
 
 
(2 intermediate revisions by one other user not shown)
Line 1: Line 1:
 +
== Problem 9==
 +
 +
When you simplify <math>\left[ \sqrt [3]{\sqrt [6]{a^9}} \right]^4\left[ \sqrt [6]{\sqrt [3]{a^9}} \right]^4</math>, the result is:
 +
 +
<math>\textbf{(A)}\ a^{16} \qquad\textbf{(B)}\ a^{12} \qquad\textbf{(C)}\ a^8 \qquad\textbf{(D)}\ a^4 \qquad\textbf{(E)}\ a^2 </math>
 +
 
== Solution ==
 
== Solution ==
 +
This simplifies to
 +
<cmath>(a^{\frac{9}{6}/3})^4 \cdot (a^{\frac{9}{3}/6})^4 = (a^{\frac{1}{2}})^4 \cdot (a^{\frac{1}{2}})^4 = (a^2)(a^2) = \boxed{a^4}</cmath>
 +
The answer is <math>\boxed{\textbf{(D)}}.</math>
 +
==See Also==
  
−
Notice that the first part and the second part of the expression are the same things, just in a different order, ie 3 and the 6 are switched around.
+
{{AHSME box|year=1956|num-b=8|num-a=10}}
−
Therefore, the question is reduced to
 
−
<cmath>2 \cdot [root{3}{root{6}{a^{9}}}]</cmath>
 
  
−
Therefore, the answer is <math>\fbox{(A) 10}</math>
+
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:29, 12 February 2021

Problem 9

When you simplify $\left[ \sqrt [3]{\sqrt [6]{a^9}} \right]^4\left[ \sqrt [6]{\sqrt [3]{a^9}} \right]^4$, the result is:

$\textbf{(A)}\ a^{16} \qquad\textbf{(B)}\ a^{12} \qquad\textbf{(C)}\ a^8 \qquad\textbf{(D)}\ a^4 \qquad\textbf{(E)}\ a^2$

Solution

This simplifies to \[(a^{\frac{9}{6}/3})^4 \cdot (a^{\frac{9}{3}/6})^4 = (a^{\frac{1}{2}})^4 \cdot (a^{\frac{1}{2}})^4 = (a^2)(a^2) = \boxed{a^4}\] The answer is $\boxed{\textbf{(D)}}.$

See Also

1956 AHSME (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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1956_AHSME_Problems/Problem_9&oldid=146266"