Question 145
$$
u_1 = a, \ u_{n+1} = u_n + f(u_n)$$
for $n = 1, 2, \dots$ with $f(x)$ is the product of all of digits of $x$. Prove that there exist $N$ such that $u_n = u_N$ for any $n \geq N$.
u_1 = a, \ u_{n+1} = u_n + f(u_n)$$
for $n = 1, 2, \dots$ with $f(x)$ is the product of all of digits of $x$. Prove that there exist $N$ such that $u_n = u_N$ for any $n \geq N$.