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
$$\frac{p-b_0}{|b_0|} = \frac{3^{c_0} \cdot b_1^*}{|b_0|}, \text{ where } 3 \nmid b_1^* \text{ and } 0 < b_1^* < \frac{p}{2}.$$
Taking $b_1 = \pm b_1^*$ such that $b_1 \equiv p \pmod{3}$, then
$$\frac{p-b_1}{|b_1|} = \frac{3^{c_1} \cdot b_2^*}{b_1^*}, \text{ where } 3 \nmid b_2^* \text{ and } 0 < b_2^* < \frac{p}{2}.$$
Taking $b_2 = \pm b_2^*$ such that $b_2 \equiv p \pmod{3}$, then
$$\frac{p-b_2}{|b_2|} = \frac{3^{c_2} \cdot b_3^*}{b_2^*}, \text{ where } 3 \nmid b_3^* \text{ and } 0 < b_3^* < \frac{p}{2}.$$
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,
$$\frac{p-b_i}{|b_i|} \cdot \frac{p-b_{i+1}}{|b_{i+1}|} \cdots \frac{p-b_{j-1}}{|b_{j-1}|} = \frac{3^{c_i} \cdot b_{i+1}^*}{b_i^*} \cdot \frac{3^{c_{i+1}} \cdot b_{i+2}^*}{b_{i+1}^*} \cdots \frac{3^{c_{j-1}} \cdot b_j^*}{b_{j-1}^*}.$$
Since $b_i = b_j$, then $b_i^* = b_j^*$. So the above expression equals to
$$3^{c_i+c_{i+1}+\cdots+c_{j-1}} = 3^n, \quad n > 0.$$
Put $b_i, b_{i+1}, \dots, b_{j-1}$ in ascending order, as desired.