The Twenty-fifth Hong Kong (China) Mathematical Olympiad

Note that $\frac{39}{1428}=\frac{13}{476}$ in lowest term. Let $$\frac{13}{476}=0\cdot b_1{b}_2\cdots b_s\overline{a_1{a}_2\cdots a_k}(1)$$ in binary decimal representation, where $a_1{a}_2\cdots a_k$ is its repetend and $k$ is its period. We can write $0.b_1{b}_2\cdots b_s=\frac{B}{2^s}$ and $0.a_1{a}_2\cdots a_k=\frac{A}{2^k}$ for some integers $A$ and $B$. Thus, $$\frac{13}{476}=\frac{B}{2^s}+\frac{A}{2^s\cdot 2^k}+\frac{A}{2^s\cdot 2^{2k}}+\frac{A}{2^s\cdot 2^{3k}}+\cdots =\frac{B}{2^s}+\frac{A}{2^s(2^k-1)}=\frac{B(2^k-1)+A}{2^s(2^k-1)}.(2)$$ Note that (2) is equivalent to (1). Therefore, it remains to find the smallest positive integer $k$ such that (2) holds for some $A,B,s$ satisfying $0\le A<2^k$ and $0\le B<2^s$. As $476=2^2\times 7\times 17$, we must have $7\mid 2^k-1$ and $17\mid 2^k-1$. Since the orders of 2 modulo 7 and 17 are 3 and 8 respectively, we have $3\mid k$ and $8\mid k$. Therefore, the smallest possible $k$ is 24. Indeed, we can take $s=2$, $B=0$ and $A=13\times \frac{2^{24}-1}{119}$, so that (2) is satisfied. This shows the period is 24.