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Poisson Games and Sudden-Death Overtime

Interview Doubler


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


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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