# 1974 AHSME Problems/Problem 28

## Problem

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 $2$,..., $a_{25}$ is $0$ or $2$? $\mathrm{(A)\ } 0\le x<1/3 \qquad \mathrm{(B) \ } 1/3\le x<2/3 \qquad \mathrm{(C) \ } 2/3\le x<1 \qquad$ $\mathrm{(D) \ } 0\le x<1/3 \text{ or }2/3\le x<1 \qquad \mathrm{(E) \ }1/2\le x\le 3/4$

## Solution

We can separate this into two cases.

Case 1: $a_1=0$ Here, clearly the minimum sum is when all of the $a_i$ are $0$, so $0\le x$. The maximum sum is when the rest of the $a_i$ are all $2$. In this case, $$\sum_{n=2}^{25}\frac{2}{3^n}<2\sum_{n=2}^{\infty}\frac{1}{3^n}=2\left(\frac{\frac{1}{9}}{1-\frac{1}{3}}\right)=\frac{1}{3}.$$ Therefore, in this case, $0\le x<1/3$.

Case 2: $a_1=2$ Here, again, the minimum sum is when the rest of the $a_i$ are all $0$, in which case the minimum sum is $\frac{2}{3}$. The maximum sum is attained when all of the $a_i$ are $2$. In that case, $$\sum_{n=1}^{25}\frac{2}{3^n}<2\sum_{n=1}^{\infty}\frac{1}{3^n}=2\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)=1.$$ Therefore, in this case, $2/3\le x<1$.

Thus, we must have $0\le x<1/3$ or $2/3\le x<1$, $\boxed{\text{D}}$.

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