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

Line 9: Line 9:
  
 
==Solution 2==
 
==Solution 2==
Substitute <math>1</math> for <math>h</math>. You get <math>1^3-(1^3-1)</math> which is 1. That is <math>h</math>, so the answer is <math>\mathrm{(C)}</math>.
+
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 ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=A|num-b=1|num-a=3}}
 
{{AMC12 box|year=2006|ab=A|num-b=1|num-a=3}}

Revision as of 18:37, 29 April 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 $h\otimes (h\otimes h)$?

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

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

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