Slovenian Mathematical Olympiad, 2008

Let us find the first few $n$. When $p=2$ we have $n=9$, when $p=3$ we have $n=49$ and when $p=5$ we have $n=513$. Now, let $p > 5$. Rewrite $n$ as $n = (p-2)(p-1)(p+1)(p+2)+9$. Since $(p-2), (p-1), p, (p+1), (p+2)$ are five consecutive positive integers, at least one of them is divisible by $5$. Since $p > 5$, $p$ is not divisible by $5$. Hence, $5$ divides one of the numbers $(p-2), (p-1), (p+1), (p+2)$ and must also divide their product $(p-2)(p-1)(p+1)(p+2)$. At least one of the numbers $p+1$ and $p+2$ is even, so this product is divisible by $2$. Thus, for $p > 5$ the number $(p-2)(p-1)(p+1)(p+2)$ is divisible by $10$. For all $p > 5$ the number $n$ has at least two digits and the final digit is $9$, so the sum of the digits is greater than $9$. We conclude that the least possible sum of the digits is $9$ and this value is attained only when $p = 2$ or $p = 5$.