Let $n=a_ka_{k-1}\ldots a_1a_0$ be a $(k+1)$ digit number.

**Step 1:** Find $a_k \times a_{k-1} \times \ldots \times a_1 \times a_0= c_tc_{t-1}\ldots c_1c_0$.

**Step 2:** Find $a_k + a_{k-1} + \ldots + a_1 + a_0= d_ld_{l-1}\ldots d_1d_0$.

Then form $n_1=c_tc_{t-1}\ldots c_1c_0d_ld_{l-1}\ldots d_1d_0$.

Repeat steps 1 and 2 for $n_1$ if $n_1$ is not a single digit number. Else, stop.

Examples: $1 \to 1; \ldots ; 9 \to 9; 10\to 01=1; 11\to 12 \to 23\to 65\to 3011\to 05=5 ; \ldots$

**Note:**
All the permutations of the digits of a number yield the same result, i.e. for instance, $127,172,217,271,712,721 \to 1410\to 06$. Hence, once all numbers from 1 to 50 have been checked, we need only check 55, 66, 77, 88 and 99 to extend our result for the range 1-100.

**Facts:**
All numbers except $38$(or $83$), $66$, and $88$ converge. $38$ and $88$ fall into the 5-cycle: $$88\to 8416 \to 14417\to 11217\to 1412 \to 88$$ while $66$ falls into the interesting 1-cycle: $$3612\to 3612$$
Also $6417$ falls into the 5-cycle: $$16818\to 38424\to 76821\to 67224\to 67221\to 16818$$

**Question:** Is the Collatz phenomenon happening here?(i.e. Do all numbers yield a sequence that converges to a single digit number or a sequence that falls into a cycle?)

**My question has been answered**. But I had some thoughts going on based on my experiment: Is the number of cycles finite?

Thanks.(MO is cool! I can't disagree.)