MHB Showcase of 2016 Consecutive Numbers w/ 100 Primes

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The discussion focuses on demonstrating the existence of 2016 consecutive integers that contain exactly 100 prime numbers. Participants explore various mathematical approaches and theorems to prove this claim. The conversation includes examples and calculations to illustrate the distribution of primes within the specified range. Key insights involve the application of number theory and prime density concepts. The thread concludes with a consensus on the feasibility of the solution presented.
kaliprasad
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Show that there exists 2016 consecutive numbers that contains exactly 100 primes.
 
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kaliprasad said:
Show that there exists 2016 consecutive numbers that contains exactly 100 primes.

My solution

we know that number of primes less than 1000 is $= 168$
Now let f(x) be number of primes in a sequence of 2016 primes starting at x.
$f(1) > 100$.
now when we move to next number the number of primes increases/decreases by 1 or remains unchanged
$f(2017!+2) = 0$ as 2016 numbers starting from this number all are composite
So from 1 going upto 2017!+2 the starting number ( $>100$) remains unchanged or increases by 1 or decreases by 1
going to 0.
Hence at some point it is 100.
 

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