Question

A positive integer is picked randomly from 10^(10^61) to 10^(10^70), inclusive.

What is the probability that it is prime?

Answer 1

but numerator is negative, denominator is positive?

Answer 2

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

Answer 3

well that is wrong! yikes

Answer 4

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

Answer 5

let me try the algebra again.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

do you know the answer?

Answer 10

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

Answer 11

how did you get it?

Answer 12

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

Answer 13

ah i see

Answer 14

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 \[\frac{1}{\ln N}\]