Link to the other post about this problem
Prove that there does not exist an integer which is doubled when the initial digit is transferred to the end.
So today I started working on this fairly nice-seeming problem. I think I have a solution but I only slept for 5 hours and it might be w*ck, since it has made me think that the question might not be very accurate (in a sense). Before I proceed, I know there's another post about this very same problem but I am providing potentially a different solution and trying to discuss whether the problem is accurate, so I hope this post isn't closed as a duplicate.
$Solution.$ Here's another attempt at finding a solution.
Let such integer be $z=10^na_n+10^{n-1}a_{n-1}+\cdots+a_0$ (decimal representation) then the integer $\tilde{z}=10^na_{n-1}+10^{n-1}a_{n-2}+\cdots+10a_0+a_n$ is obtained by moving the initial digit of $z$ to the last.
Now observe the following,
$$10z+a_n = 10^{n+1}a_n+10^{n}a_{n-1}+\cdots+10a_0+a_n$$ $$=10^{n+1}a_n+\tilde{z}=10^{n+1}a_n+2z$$
and so,
$$10(10^n-1)a_n=8z$$
and we can see that this is false since $8\not|10$ and $8\not|(10^n-1)$
At this point I ask if I made any stupid mistakes, please let me know One comment about this problem is that the integer $0$ satisfies the condition, where taking the initial digit to the last (which happens to be the same) doubles the integer? If so, should the question be modified?