## Task 26/83

How many (genuine) four-digit natural numbers are there that are divisible by 11 and whose cross sum is also divisible by 11?
A natural number is divisible by 11 if and only if its alternating sum of digits is divisible by 11. If

$$
x=1000 a_{3}+100 a_{2}+10 a_{1}+a_{0}
$$

is a number that meets the conditions stated in the problem ( $a_{i} \in \mathbb{N}$ for $i=0, 1, 2, 3$, $0 \leq a_{i} \leq 9$ for $i=0, 1, 2$, and $1 \leq a_{3} \leq 9$ ), then

$$
a_{3}-a_{2}+a_{1}-a_{0}=11 r \quad \text{and} \quad a_{3}+a_{2}+a_{1}+a_{0}=11 m
$$

with $r \in \{-1, 0, +1\}$ and $m \in \{+1, +2, +3\}$. By adding and subtracting these equations, we get

$$
a_{3}+a_{1}=5.5(m+r) \quad ; \quad a_{2}-a_{0}=5.5(m-r)
$$

From this, it immediately follows that $m \equiv r \pmod{2}$ because $a_{i} \in \mathbb{N}$. Therefore, the following pairs $(r, m)$ are possible: $(-1, 1), (-1, 3), (0, 2), (1, 1), (1, 3)$. From these pairs, we get five systems of equations:

$$
\begin{array}{cc}
a_{3}+a_{1}=0 & a_{2}+a_{0}=11 \\
a_{3}+a_{1}=11 & a_{2}+a_{0}=22 \\
a_{3}+a_{1}=11 & a_{2}+a_{0}=11 \\
a_{3}+a_{1}=11 & a_{2}+a_{0}=0 \\
a_{3}+a_{1}=22 & a_{2}+a_{0}=11
\end{array}
$$

Only the third and fourth systems have solutions within the defined range of the $a_{i}$. For $a_{3}+a_{1}=11$ to hold, $a_{3}, a_{1} > 1$ must be true.

Thus, the pairs $\left(a_{3}, a_{1}\right)=(2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3)$, and $(9, 2)$ are possible.

From the third system, the same possibilities for $\left(a_{2}, a_{0}\right)$ arise, and from the fourth system, only $a_{2}=a_{0}=0$ can follow, so that with 8 possibilities for $\left(a_{3}, a_{1}\right)$, there are nine for $\left(a_{2}, a_{0}\right)$. In total, there are $8 \cdot 9 = 72$ numbers of the required type.