Difference between revisions of "1984 AHSME Problems/Problem 11"
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==Solution== | ==Solution== | ||
− | 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} | + | 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 10:55, 21 July 2015
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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.