anonymous
  • anonymous
A positive integer is picked randomly from 10^(10^61) to 10^(10^70), inclusive. What is the probability that it is prime?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
but numerator is negative, denominator is positive?
anonymous
  • anonymous
i would not bet a stick of gum that the answer is right
anonymous
  • anonymous
well that is wrong! yikes

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anonymous
  • anonymous
\[\pi(n)\approx \frac{n}{\log(n)}\] embarrassing too let me get rid of it
anonymous
  • anonymous
let me try the algebra again.
anonymous
  • anonymous
i can't write with all these powers to powers, so lets put \[m=10^{10^{60}}\] \[n=10^{10^{70}}\]
anonymous
  • anonymous
then i guess you estimate at \[\pi(n)-\pi(m)=\frac{m\log(n)-n\log(m)}{mn}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
do you know the answer?
anonymous
  • anonymous
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
anonymous
  • anonymous
how did you get it?
anonymous
  • anonymous
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
anonymous
  • anonymous
ah i see
anonymous
  • anonymous
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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