\section*{Problem 5 - 340935}

Determine all positive integers \(n\) for which each of the six numbers

\[
n, \quad n+2, \quad n+6, \quad n+8, \quad n+12, \quad n+14
\]

is a prime number.
}

Among the numbers \(n, n+6=5+(n+1), n+2, n+8=5+(n+3)\) and \(n+14=10+(n+4)\), exactly one is divisible by 5, since exactly one of the five consecutive natural numbers from \(n\) to \(n+4\) is divisible by 5.

The six numbers can only all be prime if the 5 is among them. Since \(n>0\), this leaves only \(n\) or \(n+2\) as possibilities. However, if \(n+2=5\), then \(n+12=15\) would not be a prime number. Therefore, only \(n=5\) remains, which indeed is a solution because \(n+2=7, n+6=11, n+8=13, n+12=17\) and \(n+14=19\) are all prime. Thus, there is exactly one such \(n\), namely \(n=5\).