Silk Road Mathematics Competition

Suppose $1/(2n+1)={\dot{a}}_1{a}_2\cdots a_n{b}_1{b}_2\cdots {\dot{b}}_n$, $a_i+b_i=9$, $i=1,2,\dots ,n$, and $2n$ is its period. Since $$1+\frac{1}{10^{2n}}+\frac{1}{10^{4n}}+\cdots =\frac{1}{1-\frac{1}{10^{2n}}}=\frac{10^{2n}}{10^{2n}-1}$$ we have $$\begin{array}{rl}(1)\frac{1}{2n+1}& =\frac{10^{2n}}{10^{2n}-1}\left(\sum _{i=1}^n\frac{a_i}{10^i}+\frac{1}{10^n}\sum _{i=1}^n\frac{b_i}{10^i}\right)\\ & =\frac{10^{2n}}{10^{2n}-1}\left(\sum _{i=1}^n\frac{a_i}{10^i}+\frac{1}{10^n}\sum _{i=1}^n\frac{9-a_i}{10^i}\right)\\ & =\frac{10^{2n}}{10^{2n}-1}\left[\left(1-\frac{1}{10^n}\right)\sum _{i=1}^n\frac{a_i}{10^i}+\frac{9}{10^n}\sum _{i=1}^n\frac{1}{10^i}\right]\\ & =\frac{10^{2n}}{10^{2n}-1}\left[\left(1-\frac{1}{10^n}\right)\sum _{i=1}^n\frac{a_i}{10^i}+\frac{9}{10^n}\cdot \frac{1}{10}\cdot \frac{1-\frac{1}{10^n}}{1-\frac{1}{10}}\right]\\ & =\frac{10^{2n}}{10^n+1}\left(\frac{1}{10^n}\sum _{i=1}^n\frac{a_i}{10^i}+\frac{1}{10^{2n}}\right)=\frac{{\sum }_{i=1}^{n}10^{n-i}{a}_i+1}{10^n+1}\end{array}$$ Therefore, $2n+1\mid 10^n+1.$ By (1) and the knowledge of numbers, we guess the sufficient and necessary condition is $$(2)2n+1\mid 10^n+1\text{ and }2n+1\nmid 10^i+1\text{ for }i=1,2,\dots ,n-1.$$ In fact, since $2n$ is the period of $1/(2n+1)$, we have $$2n+1\mid 10^{2n}-1\text{ and }2n+1\nmid 10^i-1\text{ for }i=1,2,\dots ,n-1.$$ So $2n+1\nmid (10^n+1)+(10^{n+i}-1)=10^n(10^i+1)$ for $i=1,2,\dots ,n-1$. Namely, $2n+1\nmid 10^i+1$ for $i=1,2,\dots ,n-1$. The condition (2) is necessary.

Now we assume that $1/(2n+1)$ satisfies the condition (2). Let $$\frac{10^n+1}{2n+1}-1=\sum _{i=1}^{n}10^{n-i}{a}_i,$$ where $0\le a_i\le 9$, $i=1,\dots ,n$ are integers. Then $$\frac{1}{2n+1}=\frac{{\sum }_{i=1}^{n}10^{n-i}{a}_i+1}{10^n+1}.$$ Put $b_i=9-a_i$. Using (1) we have $$\frac{1}{2n+1}={\dot{a}}_1{a}_2\cdots a_n{b}_1{b}_2\cdots {\dot{b}}_n.$$ For $1\le i\le n$, because $$(10^n+1)+(10^i-1)=10^i(10^{n-i}+1),$$ so $2n+1\nmid 10^i-1$. For $n<i<2n$, because $$(10^n+1)+(10^i-1)=10^n(10^{i-n}+1),$$ we also have $2n+1\nmid 10^i-1$. Therefore the period of $1/(2n+1)$ is $2n$. That is, the condition (2) is sufficient. By the condition (2), if $1/(2n+1)$ is a fraction we look for, then $2n+1$ does not have factors 3 and 5. So the possible values of $1/(2n+1)$ are 11, 13, 17, 19, 23, .... Since $11,13\nmid 10^3+1=1001$, 11 and 13 do not satisfy the condition. Consider $17=2\times 8+1$. We have $$10^8+1=(17\times 4-2{)}^4=2^4+1=0(\mathrm{mod}17).$$ For $i=1,2,3,4$ it is obvious that $17\nmid 10^i+1$. For $i=5,6,7$, since $(10^8+1)-(10^i+1)=10^i(10^{8-i}-1)$, it is obvious too that $17\nmid 10^i+1$. Therefore, $1/17$ is a fraction we need. Similarly, consider $19=2\times 9+1$. $$10^9+1=(53\times 19-7{)}^3+1=-342=0(\mathrm{mod}19).$$ It is easy to check that $19\nmid 10^i+1$ for $1\le i<9$. So $1/19$ is another fraction we want to seek.