Suppose $x$ is an $n$ -digit number having the property that the value of $x$ is doubled if the last digit of $x$ is moved to the front.

Find the smallest possible value of $n$

The answer is 18.

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Let $b$ be the last digit of a number $10a+b$ . Then for some $x$

$10^x b + a=2(10a+b)$

which means that

$b=\dfrac{19a}{10^x-2}$

Since $b$ is a single digit, the smallest $x$ for which $10^x-2$ has $19$ as a factor is $x=17$ , so that the number is

$52631578947368420b+b$

Since $b=1$ leads to $52631578947368421$ , the double of which is $105263157894736842$ , we try the next value $b=2$ , which leads to

$105263157894736842$ , the double of which is $210526315789473684$