# 1984 AHSME Problems/Problem 11

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## 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 $\left(\left(x^{-2}\right)^{-2}\right)^{\cdots})) (\text{n times})=x^{(-2)^n}, \boxed{\text{A}}$.

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