Difference between revisions of "2006 AMC 10B Problems/Problem 2"

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== See Also ==
 
== See Also ==
*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|num-b=1|num-a=3}}
 
 
*[[2006 AMC 10B Problems/Problem 1|Previous Problem]]
 
 
 
*[[2006 AMC 10B Problems/Problem 3|Next Problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 22:52, 7 September 2011

Problem

For real numbers $x$ and $y$, define $x \spadesuit y = (x+y)(x-y)$. What is $3 \spadesuit (4 \spadesuit 5)$?

$\mathrm{(A) \ } -72\qquad \mathrm{(B) \ } -27\qquad \mathrm{(C) \ } -24\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 72$

Solution

Since $x \spadesuit y = (x+y)(x-y)$:

$3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit((4+5)(4-5)) =  3 \spadesuit (-9) = (3+(-9))(3-(-9)) = -72 \Rightarrow A$

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions
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