## Task 1 - 210931

Four pairs of statements are made about a natural number $x$:

Pair A: (1) $x$ is a two-digit number.

(2) $x$ is less than 1000.

Pair B: (1) The second digit of the number $x$ is 0.

(2) The sum of the digits of the number $x$ is 11.

Pair C: (1) $x$ is written with exactly three digits, and all three digits are the same.

(2) $x$ is divisible by 37.

Pair D: (1) The sum of the digits of the number $x$ is 27.

(2) The product of the numbers represented by the individual digits of $x$ is 0.

Investigate whether there are natural numbers $x$ with $x \neq 0$ for which in each of the four pairs A, B, C, D, one statement is true and one statement is false!

If such numbers $x$ exist, determine all these numbers!
Since every two-digit number is also less than 1000, $\mathrm{A}(1)$ cannot be true, as otherwise $\mathrm{A}(2)$ would also be true. Therefore, $x$ is not a two-digit number but is less than 1000, so it must be a one- or three-digit number.

However, $x$ cannot be a one-digit number, because then both $\mathrm{C}(1)$ ($x$ is written with three identical digits) and $\mathrm{C}(2)$ ($x$ is divisible by 37) would be false due to $x > 0$. Therefore, $x$ is a three-digit natural number.

If $\mathrm{C}(1)$ were true, there would be a digit $n$ such that $x = n \cdot 111 = n \cdot 3 \cdot 37$, which would also make $\mathrm{C}(2)$ true, contradicting the problem statement. Therefore, $x$ is divisible by 37 but not by 3. Consequently, the sum of the digits of $x$ is not divisible by 3, so it cannot be 27, making $\mathrm{D}(1)$ false and $\mathrm{D}(2)$ true. Therefore, one of the digits of $x$ must be zero. This cannot be the leading digit, as otherwise the number would not be three digits.

We now consider two cases, depending on which digit of $x$ is zero.

1. Case: The second digit (tens place) of $x$ is zero. Then $\mathrm{B}(1)$ is true and $\mathrm{B}(2)$ must be false, so $x$ is a three-digit number divisible by 37 but not by 3, with 0 in the tens place and a digit sum different from 11. The three-digit multiples of 37 are:

$111, 148, 185, 222, 259, 296, 333, 370, 407, 444, 481, 518, 555, 592, 629, 666, 703, 740, 777, 814$,

$851, 888, 925, 962$, and 999

Of these, only the numbers 407 and 703 have a tens digit of 0, but 407 has a digit sum of 11, so the only solution for this case is $x = 703$.

2. Case: The second digit of $x$ is not zero. Then the units digit of $x$ must be zero, and since $\mathrm{B}(1)$ is false, $\mathrm{B}(2)$ must be true, and the number must have a digit sum of 11. Since the units digit of $x$ is 0, $x$ is not only divisible by 37 but also by 10, and thus, because $\operatorname{gcd}(37, 10) = 1$, it is also divisible by 370.

Therefore, the only possibilities for $x$ in the considered interval are 370 and 740, with the first possibility being excluded due to its digit sum of 10. The only solution for this case is $x = 740$.

It is easily verified that both values 703 and 740 each satisfy exactly one of the statements in each pair A, B, C, and D.