# Is it possible to construct a sequence that ends in 1000000000?

Starting from the number $1$ we write down a sequence of numbers where the next number in the sequence is obtained from the previous one either by doubling it or rearranging its digits (not allowing the first digit of the rearranged number to be $0$) for instance a sequence might begin $$1,2,4,8,16,61,122,212,424,\ldots$$ Is it possible to make a sequence that ends in $1000000000$ and a sequence that ends in $9876543210$. Please show me how and if there is any working.

• 1000000000 shouldn't be hard, if you have just one non-zero digit, could it have been rearranged from any number? EDIT: oh, ok, maybe not that obvious. Aug 16, 2015 at 9:21
• Hint: $1000 = 8\cdot 125$. Aug 16, 2015 at 9:28
• You can try to work backwards: halve the number successively a few times. Rearrange. Repeat. If you're lucky you can eventually rearrange it into a power of $2$. At least I think that that method is slightly more constructive. Aug 16, 2015 at 10:17
• @DanielFischer Combine that with 8⋅625 and you will have covered all exponents except from three possibilities. 10 and 100 are impossible to reach. 100000 is possible if you go via 256, 265, 1060, 1600, 51200, 12500. Aug 16, 2015 at 13:31
• Aug 19, 2015 at 8:38

We first note that we can attain $1000$ from $1$ by $$1,2,4,8,16,32,64,128,256,512,125,250,500,1000$$ Therefore $1000^n$ can be attained by performing the same operations as above with $1$ replaced by $1000^{n-1}$.
On the other hand, note that rearranging the digits does not change the remainder when divided by $3$, and $2^n \neq 0 \pmod{3}$. Therefore the sequence would not reach any multiple of $3$.