Difference between revisions of "1961 AHSME Problems/Problem 1"
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| + | == Problem== | ||
| − | When simplified <math>(-\frac{1}{125})^{-\frac{2}{ | + | When simplified, <math>(-\frac{1}{125})^{-2/3}</math> becomes: |
| + | |||
| + | <math>\textbf{(A)}\ \frac{1}{25} \qquad | ||
| + | \textbf{(B)}\ -\frac{1}{25} \qquad | ||
| + | \textbf{(C)}\ 25\qquad | ||
| + | \textbf{(D)}\ -25\qquad | ||
| + | \textbf{(E)}\ 25\sqrt{-1}</math> | ||
| + | |||
| + | ==Solution== | ||
| + | To remove the negative exponent, flip the fraction of the base. This results in <math>(-125)^{2/3}</math>. | ||
| + | |||
| + | Then, cube root <math>-125</math> and and square the result to get the answer of <math>25</math>, or answer choice <math>\boxed{\textbf{(C)}}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1961|before=First Question|num-a=2}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 11:04, 17 May 2018
Problem
When simplified, 
becomes:
Solution
To remove the negative exponent, flip the fraction of the base. This results in 
.
Then, cube root 
and and square the result to get the answer of 
, or answer choice 
.
See Also
| 1961 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
