Difference between revisions of "1984 AHSME Problems/Problem 11"
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| − | Notice that taking the reciprocal of a number is equivalent to raising the number to the <math> -1st </math> power. Therefore, taking the reciprocal of a number <math> x </math> and then squaring it is <math> (x^{-1})^2=x^{-2} </math>. Also, since multiplication is commutative, this is equivalent to squaring the number then taking the reciprocal of it. Doing this <math> n </math> times equals <math> x^{-2^{-2^{ | + | Notice that taking the reciprocal of a number is equivalent to raising the number to the <math> -1st </math> power. Therefore, taking the reciprocal of a number <math> x </math> and then squaring it is <math> (x^{-1})^2=x^{-2} </math>. Also, since multiplication is commutative, this is equivalent to squaring the number then taking the reciprocal of it. Doing this <math> n </math> times equals <math> \left(\left(x^{-2}\right)^{-2}\right)^{\cdots}}} (\text{n times})=x^{(-2)^n}, \boxed{\text{A}} </math>. |
==See Also== | ==See Also== | ||
{{AHSME box|year=1984|num-b=10|num-a=12}} | {{AHSME box|year=1984|num-b=10|num-a=12}} | ||
Revision as of 16:55, 1 October 2011
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 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
