Answer : $n = 4, 48, 72$ [ Proof ] Let $k$ be the number of digits of $n$. We have $n = \prod_{i=1}^k p_i^{e_i} \qquad$ ( the prime factorization of $n$ ) We have $10^k > n > \prod_{i=1}^k 2^{e_i} = 2^{\sum_{i=1}^k e_i} = 2^{\sum_{i=1}^k p_i} \geqq 2^{k^2+1}$ $\therefore 10^k > 2^{k^2+1}$ So we can say that $k \leq 2$. (1) the case $k=2$ WLOG, $p_1 < p_2$. We have $2^7 > 100 > n > 2^{e_1+e_2} = 2^{p_1+p_2}$. So $p_1+p_2 \leq 6$ and then $(p_1,p_2)=(2,3)$. Hence $n \in \{2^4 3^1, 2^3 3^2\}=\{48, 72\}$ (2) the case $k=1$ We have $n = p^p < 10$. Hence $n=4$ is the only solution in this case. We get all the solutions $n \in \{4, 48, 72\}$.