1961 AMC 12 Problem 1

Attempt Problem 1 of the 1961 AMC 12 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 1961 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

When simplified, (−1125)−23\left(-\dfrac{1}{125}\right)^{-\frac{2}{3}} becomes:
125\dfrac1{25}
−125-\dfrac1{25}
2525
−25-25
25−125\sqrt{-1}
Answer: C
Concepts:exponentradicalalgebraic manipulation
Small hint:
Apply the cube root before squaring
Big hint:
The negative exponent then takes the reciprocal
Solution:
Because the cube root is real, (−1125)23=(−15)2=125. \left(-\frac1{125}\right)^{\frac{2}{3}} =\left(-\frac15\right)^2=\frac1{25}. The negative exponent takes the reciprocal, giving 25.25.
Therefore, the correct answer is C.
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