2. Let $n$ be a natural number such that the numbers $n, n+2$, and $n+4$ are prime. Determine the smallest natural number which, when divided by $n$ gives a remainder of 1, when divided by $n+2$ gives a remainder of 2, and when divided by $n+4$ gives a remainder of 6.
2. For $n>3, n=3k$ cannot be prime, $n=3k+1$ checks $n+2=3k+3$ which is not prime, being divisible by 3, and $n=3k+2$ checks $n+4=3k+6$ which is not prime, being divisible by 3. It follows that $n$ can be 2 or 3.

For $n=2$ the numbers $2,4,6$ do not all satisfy being prime.

For $n=3$ the numbers $3,5,7$ all satisfy being prime, so the numbers are $3,5,7$. $3p$

If $m=3p+1, m=5q+2, m=7s+6$ it follows that $m+8=3p+9$ is divisible by 3, $m+8=5q+10$ is divisible by 5, $m+8=7s+14$ is divisible by 7, so $m+8$ is divisible by 3, 5, 7. Since the least common multiple of 3, 5, 7 is 105, it follows that $m=97$ is the smallest number that satisfies the properties. $4p$