# 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 $\frac{1}{2}$ 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

Notice that $\log_{10}{(d+1)}-\log_{10}{d}=\log_{10}{\frac{d+1}{d}}$. Therefore, the probability of choosing $2$ is $\log_{10}{\frac{3}{2}}$. The probability that the digit is chosen out of the set is twice that, $2\log_{10}{\frac{3}{2}}=\log_{10}{\left(\frac{3}{2}\right)^2}=\log_{10}{\frac{9}{4}}$.

$\log_{10}{\frac{9}{4}}=\log_{10}{9}-\log_{10}{4}$

$=(\log_{10}{9}-\log_{10}{8})+(\log_{10}{8}-\log_{10}{7})+(\log_{10}{7}-\log_{10}{6})+(\log_{10}{6}-\log_{10}{5})+(\log_{10}{5}-\log_{10}{4})$

which is the probability that the digit is from the set $\{4, 5, 6, 7, 8\}, \boxed{\text{C}}$.