## Task $8 / 90$

Let $n$ be a natural number with exactly $m$ digits in decimal notation $(m \geq 2)$, which is divisible by 11. By reversing the sequence of digits, the natural number $n^{\prime}$ (not necessarily with exactly $m$ digits) is formed. How many sums $n+n^{\prime}$ are divisible by 11?
By reversing the sequence of digits, the alternating sum of digits changes at most by the sign. Therefore, if $n$ is divisible by 11, then $n^{\prime}$ and consequently the sum $n+n^{\prime}$ are also divisible by 11. Thus, there are as many sums $n+n^{\prime}$ that are divisible by 11 as there are $m$-digit natural numbers $n$ that are divisible by 11. For the number $l$ of such numbers, we have

$$
l=\left[\frac{10^{m}}{11}\right]-\left[\frac{10^{m-1}}{11}\right]=9 \cdot \frac{10^{m-1}+(-1)^{m}}{11}
$$

Here, $\left[\frac{a}{b}\right]$ denotes the integer part of the fraction $\frac{a}{b}$.