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anonymous

  • 4 years ago

A positive integer is picked randomly from 10^(10^61) to 10^(10^70), inclusive. What is the probability that it is prime?

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  1. anonymous
    • 4 years ago
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    but numerator is negative, denominator is positive?

  2. anonymous
    • 4 years ago
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    i would not bet a stick of gum that the answer is right

  3. anonymous
    • 4 years ago
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    well that is wrong! yikes

  4. anonymous
    • 4 years ago
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    \[\pi(n)\approx \frac{n}{\log(n)}\] embarrassing too let me get rid of it

  5. anonymous
    • 4 years ago
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    let me try the algebra again.

  6. anonymous
    • 4 years ago
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    i can't write with all these powers to powers, so lets put \[m=10^{10^{60}}\] \[n=10^{10^{70}}\]

  7. anonymous
    • 4 years ago
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    then i guess you estimate at \[\pi(n)-\pi(m)=\frac{m\log(n)-n\log(m)}{mn}\]

  8. anonymous
    • 4 years ago
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    so dividing by total number of numbers between them you get \[\frac{m\log(n)-n\log(m)}{mn(n-m)}\] but i have a feeling there is a much better and more sophisticated way to do this

  9. anonymous
    • 4 years ago
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    do you know the answer?

  10. anonymous
    • 4 years ago
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    i'd say the answer is nearly identical to 1/ln(10^(10^70)) which is less than 1 in 10^70 chance of getting a prime

  11. anonymous
    • 4 years ago
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    how did you get it?

  12. anonymous
    • 4 years ago
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    when comparing n and m, you might use the fact that n is much bigger than m. try n = 10^(69999999990000..[61 zeros]...0000) * m or something ... rewrite all the algebra so that m/n terms can be discarded when compared to non-(m/n) terms

  13. anonymous
    • 4 years ago
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    ah i see

  14. anonymous
    • 4 years ago
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    also it might be \[\pi(N) - \pi(M) \approx \frac{N}{\ln(N)} - \frac{M}{\ln(M)}\]and\[\frac{\pi(N) - \pi(M)}{N-M}.\]I would show that second equation is \[\frac{1}{\ln N}\]

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