Problem 847

Proof by contradiction. Assume that among the 20 digits of $p^{n}$, no three or more are the same, then at most two digits are the same. Thus, the 10 digits $\{0,1, 2, \cdots, 9\}$, in the 20 positions of $\boldsymbol{p}^{n}$, each digit appears exactly twice (otherwise, $p^{n}$ would have fewer than 20 digits). Therefore, the sum of the digits of $p^{n}$ is $2 \times(0 +1+2+\cdots+9)=90$. So, $p^{n}$ is divisible by 3. Thus, $p$ is divisible by 3. But it is given that $p$ is a prime number greater than 3, which is a contradiction. Therefore, $p^{n}$ must have at least three identical digits.