# 1985 AHSME Problems/Problem 24

## Problem

A non-zero digit is chosen in such a way that the probability of choosing digit $d$ is $\log_{10}{(d+1)}-\log_{10}{d}$. The probability that the digit $2$ is chosen is exactly $1/2$ the probability that the digit chosen is in the set

$\mathrm{(A)\ } \{2,3\} \qquad \mathrm{(B) \ }\{3,4\} \qquad \mathrm{(C) \ } \{4,5,6,7,8\} \qquad \mathrm{(D) \ } \{5,6,7,8,9\} \qquad \mathrm{(E) \ }\{4,5,6,7,8,9\}$

## Solution

We have $\log_{10}{(d+1)}-\log_{10}{d} = \log_{10}{\left(\frac{d+1}{d}\right)}$, so the probability of choosing $2$ is $\log_{10}{\left(\frac{3}{2}\right)}$. The probability that the digit chosen is in the set must therefore be \begin{align*}2\log_{10}{\left(\frac{3}{2}\right)} = &\log_{10}{\left(\left(\frac{3}{2}\right)^2\right)} \\ = &\log_{10}{\left(\frac{9}{4}\right)} \\ = &\log_{10}{9}-\log_{10}{4} \\ = &\left(\log_{10}{9}-\log_{10}{8}\right)+\left(\log_{10}{8}-\log_{10}{7}\right)+\left(\log_{10}{7}-\log_{10}{6}\right) \\ &+\left(\log_{10}{6}-\log_{10}{5}\right)+\left(\log_{10}{5}-\log_{10}{4}\right),\end{align*} which, by definition, is the probability that the digit chosen is in the set $\boxed{\text{(C)} \ \{4,5,6,7,8\}}$.