Difference between revisions of "1975 AHSME Problems/Problem 18"

(Created page with "== Problem 18 == A positive integer <math>N</math> with three digits in its base ten representation is chosen at random, with each three digit number having an equal chance...")
 
 
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The probability of choosing <math>128</math>, <math>256</math>, or <math>512</math> from <math>900</math> numbers is <math>\frac{3}{900} = \frac{1}{300}</math>, which gives us the answer <math>\boxed{\textbf{(D)}\ 1/300}</math>. ~[https://artofproblemsolving.com/wiki/index.php/User:Jiang147369 jiang147369]
 
The probability of choosing <math>128</math>, <math>256</math>, or <math>512</math> from <math>900</math> numbers is <math>\frac{3}{900} = \frac{1}{300}</math>, which gives us the answer <math>\boxed{\textbf{(D)}\ 1/300}</math>. ~[https://artofproblemsolving.com/wiki/index.php/User:Jiang147369 jiang147369]
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==See Also==
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{{AHSME box|year=1975|num-b=17|num-a=19}}
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{{MAA Notice}}

Latest revision as of 16:20, 19 January 2021

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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