9.5. For seven natural numbers $a, b, c, a+b-c, a+c-b, b+c-a, a+b+c$ it is known that all of them are different prime numbers. Find all values that the smallest of these seven numbers can take.
Answer: 3.

Solution. From the condition, it follows that $a, b, c$ are also primes. If the smallest of the seven numbers were equal to two, the last four numbers would be different even numbers, which means they could not all be prime. If all seven numbers are greater than three, due to their primality, they are not divisible by 3, and their remainders when divided by 3 are 1 or -1. Consider the remainders of the numbers $a, b, c$ when divided by 3. If all their remainders are equal, then the number $a+b+c$ is divisible by 3. If two of them are equal and the third has the opposite sign, then one of the numbers $a+b-c, a+c-b, b+c-a$ is divisible by 3. In both cases, one of the seven numbers is divisible by 3 and, by assumption, is greater than 3, which contradicts its primality. Therefore, the smallest of the seven numbers in the condition can only be equal to 3. An example of such numbers: $a=7, b=13, c=17, a+b-c=3, a+c-b=11, b+c-a=23, a+b+c=37$.

Grading criteria. Correct answer with an example: 2 points. Proof of the minimality of the number 3: 5 points. If the case with two is not considered - deduct 2 points.