There exists a block of 1000 consecutive positive integers containing no prime numbers, namely, $1001!+2,1001!+3,...,1001!+1001$. Does there exist a block of 1000 consecutive positive intgers containing exactly five prime numbers?
Problem
Source: ToT - 2001 Fall Senior O-Level #2
Tags: number theory, prime numbers, number theory unsolved
chronondecay
22.08.2011 02:15
Yes. Note that the set $\{1, 2, \ldots, 1000\}$ contains more than 5 primes. Now just move this interval to $\{2, 3, \ldots, 1001\}, \{3, 4, \ldots, 1002\}, \ldots$. Each move changes the number of primes by at most one. Since we start off with more than 5 primes in the interval and there is one point in which we have no primes in the interval, at some point we must have had exactly 5 primes in the interval.
nikeballa96
22.08.2011 05:54
bluecarneal wrote: There exists a block of 1000 consecutive positive integers containing no prime numbers, namely, $1001!+2, 1001!+3,..., 1001!+1001$. Does there exist a block of 1000 consecutive positive intgers containing exactly five prime numbers? (The sequence wasn't showing up before)
mavropnevma
22.08.2011 09:45
See the related http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=425665. I am pretty sure one form or another has appeared (more than once) on AoPS.