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 $y$ be the final result when one starts with a number $x\not=0$ and alternately squares and reciprocates $n$ times each. Assuming the calculator is completely accurate (e.g. no roundoff or overflow), then $y$ equals

$\mathrm{(A) \ }x^{((-2)^n)} \qquad \mathrm{(B) \ }x^{2n} \qquad \mathrm{(C) \ } x^{-2n} \qquad \mathrm{(D) \ }x^{-(2^n)} \qquad \mathrm{(E) \ } x^{((-1)^n2n)}$

Solution

Notice that taking the reciprocal of a number is equivalent to raising the number to the $-1st$ power. Therefore, taking the reciprocal of a number $x$ and then squaring it is $(x^{-1})^2=x^{-2}$ . Also, since multiplication is commutative, this is equivalent to squaring the number then taking the reciprocal of it. Doing this $n$ times equals $\left(\left(x^{-2}\right)^{-2}\right)^{\cdots})) (\text{n times})=x^{(-2)^n}, \boxed{\text{A}}$ .

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.

Page

Toolbox

Search

1984 AHSME Problems/Problem 11

Problem

Solution

See Also