1974 AMC 12 Problem 28

Question

Which of the following is satisfied by all numbers

x

of the form

x=\frac{a_1}{3}+\frac{a_2}{3^2}+\cdots+\frac{a_{25}}{3^{25}},

where

a_1

is

0

or

2,

a_2

is

0

or

2,

\ldots,

a_{25}

is

0

or

2?

0\le x\lt\frac{1}{3}

\frac{1}{3}\le x\lt\frac{2}{3}

\frac{2}{3}\le x\lt1

0\le x\lt\frac{1}{3}

or

\frac{2}{3}\le x\lt1

\frac{1}{2}\le x\le\frac{3}{4}

Separate the cases

a_1=0

and

a_1=2

Bound the remaining terms by the infinite geometric tail

\sum_{n=2}^{\infty}\frac{2}{3^n}

If

a_1=0,

then

0\le x\le\sum_{n=2}^{25}\frac2{3^n} \lt\sum_{n=2}^{\infty}\frac2{3^n}=\frac13.

If

a_1=2,

then

x\ge\frac{2}{3},

while

x\le\sum_{n=1}^{25}\frac2{3^n} \lt\sum_{n=1}^{\infty}\frac2{3^n}=1.

Hence every such

x

lies in one of the two stated outer thirds.

Therefore, the correct answer is D.

Problem 28 in other years