Pinko wrote: $a$ and $b$ are fixed real numbers. With $x_n$ we denote the sum of the digits of $an+b$ in the decimal number system. Prove that the sequence $x_n$ contains an infinite constant subsequence. I suppose, it should be read as "...With $x_n$ we denote the sum of the digits of $\lfloor an+b\rfloor $ in the decimal number system...", since otherwise $an+b$ may have infinity number of digits after the decimal dot. For any sufficiently large $n\in \mathbb{N}$ there exists $m=m(n)\in \mathbb{N}$ such that $$10^n\le am+b<10^n +a$$It is true for any $n\geq n_0$ where $10^{n_0}>a+b$. It means the sum of digits of $\lfloor am+b\rfloor$ is less than $9k+1$, where $k$ is the number of digits of $\lfloor a\rfloor$. Thus $x_{m(n)},n>n_0$ is a bounded sequence of natural numbers, hence some value is repeated infinitely many times.