1975 AHSME Problems/Problem 18

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Problem 18

A positive integer $N$ with three digits in its base ten representation is chosen at random, with each three digit number having an equal chance of being chosen. The probability that $\log_2 N$ is an integer is

$\textbf{(A)}\ 0 \qquad  \textbf{(B)}\ 3/899 \qquad  \textbf{(C)}\ 1/225 \qquad  \textbf{(D)}\ 1/300 \qquad  \textbf{(E)}\ 1/450$


Solution

In order for $\log_2 N$ to be an integer, $N$ has to be an integer power of $2$. Since $N$ is a three-digit number in decimal form, $N$ can only be $2^{7} = 128,\ 2^{8} = 256,\ \text{or}\ 2^{9} = 512$. From $100$ to $999$, there are $900$ three-digit numbers in total.

The probability of choosing $128$, $256$, or $512$ from $900$ numbers is $\frac{3}{900} = \frac{1}{300}$, which gives us the answer $\boxed{\textbf{(D)}\ 1/300}$. ~jiang147369

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