1984 AHSME Problems/Problem 11
Problem
A calculator has a key that replaces the displayed entry with its square, and another key which replaces the displayed entry with its reciprocal. Let 
be the final result when one starts with a number 
and alternately squares and reciprocates 
times each. Assuming the calculator is completely accurate (e.g. no roundoff or overflow), then equals
Solution
Notice that taking the reciprocal of a number is equivalent to raising the number to the 
power. Therefore, taking the reciprocal of a number 
and then squaring it is 
. Also, since multiplication is commutative, this is equivalent to squaring the number then taking the reciprocal of it. Doing this times equals $\left(\left(x^{-2}\right)^{-2}\right)^{\cdots}}} (\text{n times})=x^{(-2)^n}, \boxed{\text{A}}$ (Error compiling LaTeX. Unknown error_msg).
See Also
| 1984 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
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| All AHSME Problems and Solutions | ||
