a) There are $10^6$ possible combinations, so the number of moves needed is at least $10^6-1$. We shall give a construction to achieve this. In step number $n$, rotate the $k$-th disc from the right $+1$, where $k=v_{10}(n)+1$. To see that this works it is enough to show that we will never get the same two combinations. But this is clear by induction: the first $10$ combinations are different. To see that the first $10^{k+1}$ combinations are different (we consider $000000$ as the first combination here), partition them into the groups of the first $10^k$, the second $10^k$, ..., the $10$-th $10^k$ combinations. Each group shares the same digit on the $k+1$-st disc from the right, but these digits vary from group to group, so different groups have different combinations. And within one group, the combinations differ due to the induction hypothesis. b) Again $10^6-1$, because we have $10^6/2$ different combinations with an odd sum of digits. Since with each rotation, the sum of digits changes, we must use this many steps.

randomusername wrote: b) Again $10^6-1$, because we have $10^6/2$ different combinations with an odd sum of digits. Since with each rotation, the sum of digits changes, we must use this many steps. Can someone explain this part?