## Task 3 - 140913

A natural number $z$ written in the decimal system is subject to the following requirements:

(1) The cross sum of $z$ should be 11.

(2) The digits of $z$ should be pairwise different.

(3) The number $z$ should be divisible by 11.

Determine all numbers $z$ that meet requirements (1) to (3)!
The divisibility rule for the number 11 states: a number is divisible by 11 if and only if its alternating sum of digits is divisible by 11. Since the sum of the digits of the number must be exactly 11, it follows that the alternating sum of digits is 0 or 11.

Thus, the number cannot be two-digit, because if the alternating sum of digits were zero, both digits would have to be the same, which contradicts (2); and an alternating sum of 11 cannot be achieved with a two-digit number.

However, the numbers can be three-digit: precisely when the middle digit is zero, the alternating sum of digits equals the sum of digits $\Rightarrow$ rule: the first and third digits must add up to 11. This applies to 209, 308, 407, 506, 605, 704, 803, 902. According to another rule, an alternating sum of 11 cannot be achieved with a three-digit number if the sum of digits is 11.

An alternating sum of zero with a three-digit number and a sum of digits of 11 cannot occur, as the following must hold: $a_{1}+a_{2}+a_{3}=11 \Rightarrow a_{1}+a_{3}=11-a_{2}$ or $a_{1}+a_{3}=a_{2}$. This simplifies to: $11-a_{2}=a_{2} \Rightarrow a_{2}=5.5$.

For a four-digit and larger number with a sum of digits of 11, an alternating sum of 11 cannot occur, as at least one non-zero digit will have a negative sign.

For a four-digit number, the sum of digits and the alternating sum of zero generally hold: $a_{1}+a_{2}+a_{3}+a_{4}=11 \Rightarrow a_{1}+a_{3}=11-a_{2}-a_{4}$ or $a_{1}+a_{3}=a_{2}+a_{4}$. And combined: $11-a_{2}-a_{4}=a_{2}+a_{4} \Rightarrow a_{2}+a_{4}=5.5$, which cannot occur.

For a five-digit number, the sum of digits and the alternating sum of zero hold: $a_{1}+a_{2}+$ $a_{3}+a_{4}+a_{5}=11 \Rightarrow a_{1}+a_{3}+a_{5}=11-a_{2}-a_{4}$ or $a_{1}+a_{3}+a_{5}=a_{2}+a_{4}$. And combined: $11-a_{2}-a_{4}=a_{2}+a_{4} \Rightarrow a_{2}+a_{4}=5.5$, which cannot occur.

The number cannot be six-digit or larger, as with 6 different digits, the minimum sum of digits $0+1+2+3+4+5=15$ exceeds the value 11.

Thus, the only solutions are the three-digit numbers as mentioned above.