Concatenated Primes!

Let's use functions $\pi\left(n\right)$ (outputs the number of primes less than or equal to $n$ ) and $p\left(n\right)$ (outputs the $n^{th}$ prime).

Since $\pi\left(10\right) = 4$ , $\pi\left(100\right) = 25$ , $\pi\left(1000\right) = 168$ and $\pi\left(10000\right) = 1229$ , we know that the prime with the $1000^{th}$ digit has $4$ digits.

Since $1000 - 1\cdot4 + 2\left(25-4\right) + 3\left(168-25\right) = 525 = 4\cdot131 + 1$ we know that the $1000^{th}$ digit is the first digit of $168 + 132 = 300^{th}$ prime.

Since, $p\left(300\right) = 1987$ we know that the answer is $\boxed{1}$ .

I was too lazy to write a program to do this for me, I only needed to check some of these values online and perform some calculations to be able to solve this. :D

Slick solution on Haskell:

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primes = sieve [2..] where
    sieve (p:xs) = p:sieve [x | x <- xs, rem x p > 0]

primestring = concat . map show $ primes
main = do
    putStrLn [primestring !! 999]