Showcase of 2016 Consecutive Numbers w/ 100 Primes

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This discussion presents a mathematical proof demonstrating the existence of 2016 consecutive integers that contain exactly 100 prime numbers. The proof utilizes the properties of prime distribution and employs the Sieve of Eratosthenes for identifying primes within the specified range. The conclusion confirms that such a sequence can be constructed, providing a significant insight into prime number theory.

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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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