3. An arithmetic progression with 10 terms has a difference $d>0$. If the absolute value of each term in the progression is a prime number, determine the smallest possible value of $d$.
Solution. If $d$ is odd, then there is at least one even term in every two consecutive terms. Since there are at least five even terms in the sequence, at least one of them is not equal to -2 and 2. Given that the absolute values of the elements in the sequence are prime numbers, then $d$ must be even. A similar discussion shows that $d$ must be divisible by 3.

We will prove that $d$ must be divisible by 5. If $d$ is not divisible by 5, then among five consecutive terms, there must be one divisible by 5, meaning there must be two numbers in the entire sequence that are divisible by 5. If the absolute value of each term in the sequence is a prime number, then these two numbers that are divisible by 5 must be 5 and -5. However, this is not possible, because the four terms between them would be $-3, -1, 1, 3$, which contradicts the condition that the absolute values of the terms in the sequence must be prime numbers. Therefore, the smallest possible value for $d$ is 30.

Indeed, $d=30$ is possible. An example of such an arithmetic sequence is

$$
-113, -83, -53, -23, 7, 37, 67, 97, 127, 157
$$