11. An $n$-digit number $x$ has the following property: if the last digit of $x$ is moved to the front, the result is $2 x$. Find the smallest possible value of $n$.
(2 marks)
某 $n$ 位數 $x$ 具以下性質: 若把 $x$ 的最後一位數字移到最前, 所得的數是 $2 x$ 。
求 $n$ 的最小可能值。
11. 18
11. Let $d$ be the last digit of $x$, and write $x=10 c+d$. After moving the last digit to the front, the number becomes $10^{n-1} d+c$. According to the question, we have $10^{n-1} d+c=2(10 c+d)$, or $\left(10^{n-1}-2\right) d=19 c$. Computing $10^{k}$ modulo 19 for $k=1,2,3, \ldots$, we get $10,5,12,6,3,11,15$, $17,18,9,14,7,13,16,8,4,2$. Hence the smallest possible value of $n-1$ is 17 . It remains to show that there is a 18 -digit number with the given property. Since $c$ is to be a 17 -digit number, from $\left(10^{n-1}-2\right) d=19 c$, we see that we should set $d=2$ when $n=18$. This corresponds to $c=10526315789473684$ (hence $x=105263157894736842$ ). It follows that the answer is 18 .