1984 AMC 12 Problem 11

Question

A calculator has a key which replaces the displayed entry with its square, and another key which replaces the displayed entry with its reciprocal. Let

y

be the final result if one starts with an entry

x\ne0

and alternately squares and reciprocates

n

times each. Assuming the calculator is completely accurate (e.g., no roundoff or overflow), then

y

equals

x^{(-2)^n}

x^{2n}

x^{-2n}

x^{-2^n}

x^{(-1)^n2n}

Answer: A

Concepts:exponent process simulation pattern recognition

Difficulty rating: 1750

Small hint:

Track only the exponent of

x

Big hint:

One squaring-reciprocating pair multiplies the exponent by

-2

Solution:

Write the displayed value as

x^e.

Squaring changes

e

to

2e,

and then reciprocating changes it to

-2e.

Starting from

e=1,

after

n

such pairs the exponent is

(-2)^n.

Thus

y=x^{(-2)^n}.

Therefore, the correct answer is A.