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

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.

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

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

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

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

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

@@ Line 5: / Line 5: @@
 <math>\textbf{(A)}\ -h\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ h\qquad\textbf{(D)}\ 2h\qquad\textbf{(E)}\ h^3</math>
-== Solution ==
+== Solution 1 ==
-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 <math>\mathrm{(C)}</math>.
+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)=\boxed{\textbf{(C) }h}.</math>
-==Solution 2==
-Substitute <math>1</math> for <math>h</math>. You get <math>1^3-(1^3-1)</math> which is <math>1</math>. That is <math>h</math>, so the answer is <math>\mathrm{(C)}</math>.
-~dragoon also aop is a god
 == See also ==
 {{AMC12 box|year=2006|ab=A|num-b=1|num-a=3}}