I assume $ a_{n + 1} = a_n + k\pi(a_n)$, or something similar? EDIT Correction has been done!

Thank you, in the original problem there was an "$ x$" so I made the mistake when translating

lambruscokid wrote: Let $ a$ and $ k$ be positive integers. Let $ a_i$ be the sequence defined by $ a_1 = a$ and $ a_{n + 1} = a_n + k\pi(a_n)$ where $ \pi(x)$ is the product of the digits of $ x$ (written in base ten) Prove that we can choose $ a$ and $ k$ such that the infinite sequence $ a_i$ contains exactly $ 100$ distinct terms Consider for example : $ a = 555555881111111111111111$ composed with $ 55555588$ followed with $ 4p$ numbers $ 1$ (here $ p = 4$) $ k = 14000014$. composed with $ 14$ followed with $ 2p - 4$ zeroes and $ 14$ again. Then : $ a_1 = 555555881111111111111111$ and so $ \pi(a_1)=10^6$ $ a_2 = 555555881125111125111111$ and so $ \pi(a_2)=10^8$ $ a_3 = 555555882525112525111111$ and so $ \pi(a_3)=10^{10}$ $ a_4 = 555556022525252525111111$ and so $ \pi(a_4)=0$ And $ a_n = a_4$ $ \forall n > 4$ We have exactly $ 4$ distinct numbers in the sequence. And it's easy to show that starting with : $ a = 55555588\cdot 10^{4p} + \frac {10^{4p} - 1}9$ $ k = 14(10^{2p - 2} + 1)$ we get exactly $ p$ distinct numbers in the sequence. You just then have to choose $ p = 100$