6. (48th Slovenian Mathematical Olympiad) Find all prime numbers $p$ such that $p+28$ and $p+56$ are also prime.
6. When 28 and 56 are divided by 3, the remainders are 1 and 2, respectively. Therefore, if the prime number \( p \) is divided by 3, the remainder is neither 1 nor 2; otherwise, one of the numbers \( p+28 \) or \( p+56 \) will be divisible by 3. However, the only prime number divisible by 3 is 3 itself. In this case, \( p+28=31 \) and \( p+56=59 \), both of which are prime numbers. This is the solution required.