SAUDI ARABIAN IMO Booklet 2023

Let $u_1=a$ be a positive integer. Define $u_{n+1}=u_n+f(u_n)$, where $f(x)$ is the product of all digits of $x$.

Observe that if $u_n$ contains a digit $0$, then $f(u_n)=0$, so $u_{n+1}=u_n$. Thus, the sequence becomes constant from that point onward.

Suppose $u_n$ does not contain any digit $0$. Then $f(u_n)\ge 1$, so $u_{n+1}>u_n$. However, as $u_n$ increases, eventually $u_n$ will contain a digit $0$ (since for any fixed $a$, the sequence increases and eventually reaches a number with a $0$ digit).

Once $u_n$ contains a $0$ digit, $f(u_n)=0$, so $u_{n+1}=u_n$, and the sequence remains constant.

Therefore, there exists $N$ such that $u_n=u_N$ for all $n\ge N$.