## 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?

1. anonymous

but numerator is negative, denominator is positive?

2. anonymous

i would not bet a stick of gum that the answer is right

3. anonymous

well that is wrong! yikes

4. anonymous

$\pi(n)\approx \frac{n}{\log(n)}$ embarrassing too let me get rid of it

5. anonymous

let me try the algebra again.

6. anonymous

i can't write with all these powers to powers, so lets put $m=10^{10^{60}}$ $n=10^{10^{70}}$

7. anonymous

then i guess you estimate at $\pi(n)-\pi(m)=\frac{m\log(n)-n\log(m)}{mn}$

8. 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

9. anonymous

10. 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

11. anonymous

how did you get it?

12. 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

13. anonymous

ah i see

14. 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 $\frac{1}{\ln N}$