Junior Macedonian Mathematical Olympiad

Let $n = p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$. From the condition of the problem
$$p_1 + p_2 + \ldots + p_k = a_1 + a_2 + \ldots + a_k.$$
We discuss the number of ciphers of the number $n$. If $n$ has 4 ciphers, then he has 4 different prime divisors. Then $n \ge 2^{14} \cdot 3 \cdot 5 \cdot 7 > 10^4$ which is not possible. If $n$ has $k > 4$ ciphers, then
$$n \ge 2^{22+3+5+7+\ldots+p_5+\ldots+p_k-(k-1)} \cdot 3 \cdot 5 \cdot 7 \cdot p_5 \dots p_k = 2^{14} \cdot 3 \cdot 5 \cdot 7 \cdot 2^{p_5+\ldots+p_k-(k-1)} \cdot p_5 \dots p_k \\ > 10^4 \cdot 10^{k-1} = 10^k$$
which again is not possible.
So, we get that $n$ has at least three ciphers.
Let $n$ have three ciphers. Then $n = p_1^{a_1}p_2^{a_2}p_3^{a_3}$. If $5|n$, then $n \ge 2^8 \cdot 3 \cdot 5 > 10^3$.
We get that the prime divisors of the number $n$ are $\le 3$. But, prime numbers $\le 3$ are 2 and 3, and in the factorization of the number $n$ there are 3 prime numbers, which is a contradiction.
Let $n$ has two ciphers. Then $n = p_1^{a_1}p_2^{a_2}$. If $5|n$, then $n \ge 2^6 \cdot 5 > 10^2$. Remains $n = 2^{a_1}3^{a_2}$ where $a_1 + a_2 = 5$. With direct checking we get that $n = 2^4 \cdot 3 = 48, n = 2^3 \cdot 3^2 = 72$ are solutions of the problem.
Let $n$ has one cipher. Then only $n = 2^2$ fulfils the condition of the problem.