Answer: 2, 13, 19 (in any order).

Solution. If the second statement of Anton is true, then both of Nastya's statements are false. Therefore, \(a + b + c = 34\). Thus, the second statement of Nastya is true. Since the sum of three prime numbers is 34, they cannot all be odd, and one of them must be 2. Therefore, the first statement of Boris is true.

Let \(a = 2\). Then \(b + c = 32\).

We can now list all pairs of prime numbers that sum to 32 and check if they satisfy \(ab + bc + ac = 311\).

However, we can proceed as follows:

\[
311 = ab + bc + ac = a(b + c) + bc = 64 + bc \text{, hence } bc = 247 \text{.}
\]

Since \(247 = 19 \cdot 13\), we get \(b = 13, c = 19\) (or vice versa).

## Grading Criteria.

- Any complete and correct solution - 7 points.
- Correct reasoning about which statements are true and which are false, but the numbers are not found or are found incorrectly - 3 points.
- The correct answer is provided with a check that it satisfies all conditions of the problem, but without proof that there are no other solutions, 2 points.
- Correctly identified 2 out of 3 true statements - 1 point.
- Only the answer is provided - 0 points.
4. (7 points) Three schoolchildren made two statements about natural numbers $a, b, c$:

Anton: 1) $a+b+c=34$
2) $a b c=56$

Boris: 1) $a b+b c+a c=311$

2) the smallest of the numbers is 5;

Nastya: 1) $a=b=c$

2) the numbers $a, b$ and $c$ are prime.

Each schoolchild has one true statement and one false statement. Find the numbers $a, b, c$.