2000 AMC 8 Problem 17

Attempt Problem 17 of the 2000 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 8 solutions, or check the answer key.

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17.

The operation ⊗\otimes is defined for all nonzero numbers by a⊗b=a2b.a\otimes b =\dfrac{a^{2}}{b}.

Determine [(1⊗2)⊗3]−[1⊗(2⊗3)].[(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)].

−23-\dfrac{2}{3}

−14-\dfrac{1}{4}

00

14\dfrac{1}{4}

23\dfrac{2}{3}

Answer: A
Concepts:custom operationfraction
Difficulty rating: 1410
Small Hint:

Evaluate each ⊗\otimes expression from the inside out

Big Hint:

Keep the parentheses; the operation is not associative

Solution:

We can calculate it as follows. [(1⊗2)⊗3]−[1⊗(2⊗3)]=[122⊗3]−[1⊗223]=[12⊗3]−[1⊗43]=(12)23−12(43)=14⋅13−34=112−34=−23. \begin{gather*} [(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)] \\ = [\dfrac{1^2}{2} \otimes 3] - [1 \otimes \dfrac{2^2}{3}] \\ = [\dfrac{1}{2} \otimes 3] - [1 \otimes \dfrac{4}{3}] \\ = \dfrac{(\frac{1}{2})^2}{3} - \dfrac{1^2}{(\frac{4}{3})} \\ = \dfrac{1}{4} \cdot \dfrac{1}{3} - \dfrac{3}{4} \\ = \dfrac{1}{12} - \dfrac{3}{4} \\= -\dfrac{2}{3}. \end{gather*}

Thus, A is the correct answer.

Problem 16#16
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