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