# 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

## See Also

 1975 AHSME (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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