Interview Doubler

Problem:

Find the smallest positive number that doubles when you move the last digit to the front.

Solution:

The answer is $105263157894736842$, and the solution is similar to Twisty Temperature. Let this number be $$x = x_{n}\cdot 10^{n-1} + ... + x_{1}\cdot 10^1 + x_{0}$$ with $0 < x_{n} < 10$, then after moving the last digit to the front we get $$y = x_{0}\cdot 10^{n-1} + (x-x_{0})/10.$$ We also have that $y = 2 x$, so our equation becomes $$19\cdot x = x_0 \cdot (10^{n}-1).$$ Now 10 is a primitive root modulo 19, and so it follows that $x$ is integral if and only if $n$ is of the form $18\cdot m$. Also note that we need $2 \le x_0 \le 9$ since we require $0< x_{n} < 10$. When $m=1$ and $x_0 = 2$ we get $x  =105263157894736842$; when $m=2$ and $x_0 = 2$ we get $$2 (10^{36}-1)/19 = 105263157894736842105263157894736842.$$ That this number indeed doubles when you move the last digit to the front can be verified by WolframAlpha.