Problem 990

Solution. Let the first positive element of the arithmetic sequence be $x$, and its difference be $d$. Denote by $n$ the number of digits in the number $x$. Now consider the next two elements of the sequence: $x+10^{n} \cdot d$ and $x+10^{n+1} \cdot d$. Let these be denoted by $b_{1}$ and $b_{2}$, respectively.

The decimal representation of $b_{1}$ can be obtained by first writing down $d$, then appending $n$ zeros, and finally adding $x$. Since $x$ is exactly $n$ digits long, when performing the addition, only $x$ needs to be written in place of the last $n$ zeros of $10^{n} \cdot d$. Thus, $x+10^{n} \cdot d$ can be obtained by writing $x$ after the number $d$.

By similar reasoning, it can be seen that $x+10^{n+1} \cdot d$ differs from this only in that a zero must be inserted between the numbers $d$ and $x$. However, it is clear from this that the sum of the digits of $b_{1}$ and $b_{2}$ is the same.

Based on the work of Sikolya Edit (Szentendre, Ferences Gymn., I. o. t.)