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.

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