\section*{Problem 2 - 171222}

Four students, A, B, C, and D, each make three statements about a natural number \(x\). Student A makes exactly two true statements, while students B, C, and D each make at least one and at most two true statements.

Determine all natural numbers \(x\) that satisfy these conditions:

(A1) \(x\) is a three-digit number.

(A2) \(500 < x < 600\).

(A3) Each of the digits 1, 3, 5, 7, 9 appears exactly once either in the decimal representation of \(x\) or in the decimal representation of the sum of the digits of \(x\); no other digits appear in either representation.

(B1) In the decimal representation of \(x\), the number of tens is the arithmetic mean of the number of hundreds and the number of units.

(B2) \(x\) is the product of three distinct prime numbers.

(B3) \(x\) is divisible by 5.

(C1) \(x\) is a perfect square.

(C2) If the hundreds digit is removed from the decimal representation of \(x\) and appended as the (new) last digit, the resulting number is a prime number.

(C3) The decimal representation of \(x\) contains at least three identical digits.

(D1) \(x\) is the product of two two-digit numbers.

(D2) \(x\) is a prime number.

(D3) \(x\) is odd.
}

Assume a number satisfies the conditions of the problem. Therefore,

(1) A makes exactly two true statements.

(2) B, C, and D each make exactly one or exactly two true statements.

If A1 were false, then A2 would also be false, contradicting (1). Therefore, A1 is true. (3) In the following, let \(h\) be the hundreds digit of the number, \(z\) the tens digit, and \(e\) the units digit.

Case 1)

Assume A3 is true, then the number is three-digit and its digit sum must be two-digit. Since the digit sum of a three-digit number is at most 27, the first digit of the digit sum must be 1 according to A3.

Let \(r\) be the units digit of the digit sum, then according to A3:

\[
h+z+e=10+r \Leftrightarrow h+z+e+r=10+2r
\]

\(h, z, e\) and \(r\) are the four digits 3, 5, 7, 9. Therefore, \(10+2r=h+z+e+r=24\). This implies \(r=7\). \(h, z\), and \(e\) are thus 3, 5, and 9.

Since A2 is not satisfied due to (1), the four combinations 359, 395, 935, and 953 are possible.

For none of the numbers 359, 395, 935, and 953 is B1 satisfied. For 359 and 953 (both are prime numbers), neither B2 nor B3 is satisfied. This contradicts (2).

The prime factorizations of the remaining two numbers are: \(395=5 \cdot 79\) and \(935=5 \cdot 11 \cdot 17\). Therefore, B2 is only satisfied for 935 and B3 for 395 and 935.

C1 and C3 do not hold for 395 or 935. Therefore, C2 must hold according to (2).

If the procedure described in C2 is carried out, the numbers 953 and 359 are obtained. Both are (as mentioned above) prime numbers, so C2 is satisfied.

For both 395 and 935, D3 is true and D2 is false. Regardless of D1, the condition in (2) is thus satisfied.

In Case 1), only the numbers 395 and 935 are possible.

Case 2)

Assume A3 is false, then the number must be between 500 and 600 due to (1) and A2.

C2 is then not satisfied, because the newly formed number would end in 5 and thus be divisible by 5. Therefore, either C1 or C3 must hold according to (2).

C1 holds only for \(23^2=529\) and \(24^2=576\) because \(22^2=484<500<600<625=25^2\). C3 holds only for 555.

For the three numbers 529, 576, and 555:

B1 is only satisfied for 555.

B2 is only satisfied for 555 \(-529=23 \cdot 23, 576=2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3, 555=3 \cdot 5 \cdot 37\).

B3 is only satisfied for 555.

All three numbers thus contradict the condition (2).

Conclusion: Only the numbers 395 and 935 are possible.

For both numbers, (1) and (2) are satisfied. The true statements are:

395: A1, A3, B3, C2, D3

935: A1, A3, B3, C2, D1, D3