Cycling Digits puzzle I'm trying to answer the following:
"I have in mind a number which, when you remove the units digit and place it at the front, gives the same result as multiplying the original number by $2$. Am I telling the truth?"
I think the answer to that is no. It's easy to prove that it's false for numbers with two digits: Let $N = d_0 + 10 \cdot d_1$. Then $2N = 2 d_0 + 20 d_1$ and the "swapped" number is $N^\prime = d_1 + 10 d_0$. We would like to have $2d_0 + 20 d_1 = d_1 + 10d_0$ which amounts to $8d_0 = 19d_1$. The smallest value for which this equality is fulfilled is $d_0 = 19, d_1 = 8$ but $19$ is not $\leq 9$ that is, is not a digit, hence there is no solution.
Using the same argument I can show that the claim is false for $3$-digit numbers. 
I conjecture that it's false for all numbers. How can I show that? Is there a more general argument than mine, for all numbers? Thanks for helps.

If you follow your argument but let $N=a+10b$ where $a$ is a single digit but let $b$ have $n$ digits, then $2N=10^na+b$ and you get $b=\frac {10^n-2}{19}a$  If $n=17$, this is integral.  Then $a$ has to be at least $2$ to make $b$ have $17$ digits.  The smallest solution is $105263157894736842$
Another way to get there is to just start multiplying.  If you guess that the ones digit is $2$, you double it the ones digit of the product will be $4$, which will be the tens digit of the first number, and so on.  Stop when the product starts with a $2$ and doesn't carry.  You get $$105263157894736842 \\ \underline {\times\ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ 2} \\ 210526315789473684$$  If you had started with a $1$, you would miss the leading zero.
