Prime Quartests - Remainder 15

If  
\[n, \; n+2, \; n+6, \; n+8\]
  are prime numbers( a prime quartet), then the remainder when  
\[n\]
  is divided by 15 is 11.
To show this consider the consecutive odd integers  
\[n, \; n+2, \; n+4, \; n+6, \; n+8\]
. At least One of these integers is divisible by 3 and one is divisible by 5, but  
\[n, \; n+2, \; n+6, \; n+8\]
  are all prime, so  
\[n+4\]
  must be divisible by 3 and also 5, so divisible by 15, hence the remainder when  
\[n\]
  is 11.

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