China Girls' Mathematical Olympiad

Proof By the Division Algorithm, there exist unique integers $q$ and $r$ such that $p=3q+r$, where $0<r<3$. Taking $b_0=r$, then $$\frac{p-b_0}{|b_0|}=\frac{3^{c_0}\cdot b_1^{\ast }}{|b_0|},\text{ where }3\nmid b_1^{\ast }\text{ and }0<b_1^{\ast }<\frac{p}{2}.$$ Taking $b_1=\pm b_1^{\ast }$ such that $b_1\equiv p(\mathrm{mod}3)$, then $$\frac{p-b_1}{|b_1|}=\frac{3^{c_1}\cdot b_2^{\ast }}{b_1^{\ast }},\text{ where }3\nmid b_2^{\ast }\text{ and }0<b_2^{\ast }<\frac{p}{2}.$$ Taking $b_2=\pm b_2^{\ast }$ such that $b_2\equiv p(\mathrm{mod}3)$, then $$\frac{p-b_2}{|b_2|}=\frac{3^{c_2}\cdot b_3^{\ast }}{b_2^{\ast }},\text{ where }3\nmid b_3^{\ast }\text{ and }0<b_3^{\ast }<\frac{p}{2}.$$ Repeating this process, we get $b_0,b_1,\dots ,b_p$. Since these $p+1$ integers are in the interval $(-\frac{p}{2},\frac{p}{2})$, a certain integer occurs twice. Suppose $b_i=b_j$, $i<j$, and $b_i$, $b_{i+1}$, ..., $b_{j-1}$ are distinct. So, $$\frac{p-b_i}{|b_i|}\cdot \frac{p-b_{i+1}}{|b_{i+1}|}\cdots \frac{p-b_{j-1}}{|b_{j-1}|}=\frac{3^{c_i}\cdot b_{i+1}^{\ast }}{b_i^{\ast }}\cdot \frac{3^{c_{i+1}}\cdot b_{i+2}^{\ast }}{b_{i+1}^{\ast }}\cdots \frac{3^{c_{j-1}}\cdot b_j^{\ast }}{b_{j-1}^{\ast }}.$$ Since $b_i=b_j$, then $b_i^{\ast }=b_j^{\ast }$. So the above expression equals to $$3^{c_i+c_{i+1}+\cdots +c_{j-1}}=3^n,n>0.$$ Put $b_i,b_{i+1},\dots ,b_{j-1}$ in ascending order, as desired.