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2004 AMC 10A Problem 2

Problem 2 of 25EasierAlgebraArithmetic

For any three real numbers a,a, b,b, and c,c, with b≠c,b \neq c, the operation ⋄\diamond is defined by ⋄(a,b,c)=ab−c.\diamond(a, b, c) = \dfrac{a}{b - c}. What is ⋄(⋄(1,2,3),⋄(2,3,1),⋄(3,1,2))?\diamond(\diamond(1, 2, 3), \diamond(2, 3, 1), \diamond(3, 1, 2))?

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Solution

The inner values are ⋄(1,2,3)=12−3=−1,⋄(2,3,1)=23−1=1,⋄(3,1,2)=31−2=−3. \begin{aligned} \diamond(1,2,3) &= \dfrac{1}{2-3} = -1, \\ \diamond(2,3,1) &= \dfrac{2}{3-1} = 1, \\ \diamond(3,1,2) &= \dfrac{3}{1-2} = -3. \end{aligned} Therefore ⋄(−1,1,−3)=−11−(−3)=−14. \begin{aligned} \diamond(-1, 1, -3) &= \dfrac{-1}{1 - (-3)} \\ &= -\dfrac{1}{4}. \end{aligned} Thus, the correct answer is B.
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