2006 AMC 12A Problems/Problem 2: Difference between revisions

Revision as of 16:26, 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.

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

Substitute $1$ for $h$ . You get $1^3-(1^3-1)$ which is $1$ . Since $h=1$ , the answer is $\boxed{\textbf{(C) }h}.$

~dragoon also aop is a god

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 11: / Line 11: @@
 ==Solution 2==
-Substitute <math>1</math> for <math>h</math>. You get <math>1^3-(1^3-1)</math> which is <math>1</math>. Since h=1, the answer is <math>\boxed{\textbf{(C) }h}.</math>
+Substitute <math>1</math> for <math>h</math>. You get <math>1^3-(1^3-1)</math> which is <math>1</math>. Since <math>h=1</math>, the answer is <math>\boxed{\textbf{(C) }h}.</math>
 ~dragoon also aop is a god