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

(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>?
  
<math>\mathrm{(A)}\ -h\qquad\mathrm{(B)}\ 0\qquad\mathrm{(C)}\ h\qquad\mathrm{(D)}\ 2h\qquad\mathrm{(E)}\ h^3</math>
+
<math>\textbf{(A)}\ -h\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ h\qquad\textbf{(D)}\ 2h\qquad\textbf{(E)}\ h^3</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 16:21, 16 December 2021

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

By the definition of $\otimes$, we have $h\otimes h=h^{3}-h$. Then $h\otimes (h\otimes h)=h\otimes (h^{3}-h)=h^{3}-(h^{3}-h)=h$. The answer is $\mathrm{(C)}$.

Solution 2

Substitute $1$ for $h$. You get $1^3-(1^3-1)$ which is $1$. That is $h$, so the answer is $\mathrm{(C)}$. ~dragoon also aop is a god

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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