\[n, \; n+2, \; n+6, \; n+8\]

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.