2006 AMC 10B Problems/Problem 2

Revision as of 19:50, 13 July 2006 by Xantos C. Guin (talk | contribs) (Added problem and solution)

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

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

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