## Task $3 / 68$

Determine six prime numbers such that they form an arithmetic sequence and their sum is a minimum.
Let $p_{1}$ be the smallest of the six prime numbers, $p_{2}, \ldots, p_{6}$ the remaining ones, and $d$ the (by definition constant) difference of the arithmetic sequence. Then we have

$$
p_{1}=p_{1}, p_{2}=p_{1}+d, p_{3}=p_{1}+2 d, p_{4}=p_{1}+3 d, p_{5}=p_{1}+4 d, p_{6}=p_{1}+5 d
$$

For $S=6 p_{1}+15 d$ to be minimized, $p_{1}$ and $d$ must be chosen as small as possible.

The values $2, 3, 5$ are not possible for $p_{1}$, because if $p_{1}=2$, then $p_{3}=2+2 d$ would not be a prime number, if $p_{1}=3$ then $p_{4}=3+3 d$ would not be a prime number, and similarly, for $p_{1}=5$, $p_{6}=5+5 d$ would not be a prime number.

This implies $p_{1} \geq 7$, the smallest possible value for $p_{1}$ is 7.

For the choice of $d$, the following considerations apply:

1. $d$ must be divisible by 2, otherwise $p_{2}=p_{1}+d$ would be even and thus not a prime number.
2. $d$ must be divisible by 3, otherwise at least one of the numbers $p_{2}, \ldots, p_{6}$ would be divisible by 3.

Proof: If $d$ is not divisible by 3, it leaves a remainder of 1 or 2 when divided by 3.

The number $p_{1}$, being a prime number, is not divisible by 3 and thus also leaves a remainder of 1 or 2.

As one can easily verify, each of the four possible combinations leads to at least one number $p_{i}$ with $i=2, \ldots, 6$ that is divisible by 3.

3. $d$ must be divisible by 5 (analogous reasoning).

Thus, $d$ is divisible by $2 \cdot 3 \cdot 5=30$, the smallest possible $d$ is therefore 30.

Indeed, the numbers $7, 37, 67, 97, 127, 157$ form a prime number sequence.

Their sum is 492; it is the smallest sum of six prime numbers that form an arithmetic sequence.