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Difference between revisions of "2006 AMC 12A Problems/Problem 2"

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(Removed bogus answer. h=1 is a bad choice of substitution since you can't distinguish between h, or h^3.)
 
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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #2]] and [[2006 AMC 10A Problems/Problem 2|2006 AMC 10A #2]]}}
 
== Problem ==
 
== Problem ==
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Define <math>x\otimes y=x^3-y</math>. What is <math>h\otimes (h\otimes h)</math>?
 
Define <math>x\otimes y=x^3-y</math>. What is <math>h\otimes (h\otimes h)</math>?
  
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<math> \mathrm{(A) \ } -h\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } h\qquad \mathrm{(D) \ } 2h</math>
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<math>\textbf{(A)}\ -h\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ h\qquad\textbf{(D)}\ 2h\qquad\textbf{(E)}\ h^3</math>
  
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<math>\mathrm{(E) \ }  h^3</math>
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== Solution 1 ==
 +
By the definition of <math>\otimes</math>, we have <math>h\otimes h=h^{3}-h</math>.
  
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== Solution ==
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Then, <math>h\otimes (h\otimes h)=h\otimes (h^{3}-h)=h^{3}-(h^{3}-h)=\boxed{\textbf{(C) }h}.</math>
  
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By the definition of <math>\otimes</math>, we have <math>h\otimes h=h^{3}-h</math>.  Then <math>h\otimes (h\otimes h)=h\otimes (h^{3}-h)=h^{3}-(h^{3}-h)=h</math>.  The answer is C.
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== See also ==
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{{AMC12 box|year=2006|ab=A|num-b=1|num-a=3}}
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{{AMC10 box|year=2006|ab=A|num-b=1|num-a=3}}
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{{MAA Notice}}
  
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== See also ==
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[[Category:Introductory Algebra Problems]]
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* [[2006 AMC 12A Problems]]
 

Latest revision as of 13:03, 27 July 2023

The following problem is from both the 2006 AMC 12A #2 and 2006 AMC 10A #2, so both problems redirect to this page.

Problem

Define $x\otimes y=x^3-y$. What is $h\otimes (h\otimes h)$?

$\textbf{(A)}\ -h\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ h\qquad\textbf{(D)}\ 2h\qquad\textbf{(E)}\ h^3$

Solution 1

By the definition of $\otimes$, we have $h\otimes h=h^{3}-h$.

Then, $h\otimes (h\otimes h)=h\otimes (h^{3}-h)=h^{3}-(h^{3}-h)=\boxed{\textbf{(C) }h}.$

See also

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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 12 Problems and Solutions
2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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. AMC logo.png

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