3. Find the smallest natural number that has exactly 70 natural divisors (including 1 and the number itself).

$(16$ points $)$
Solution: Let $n$ be the required natural number, $n=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{m}^{k_{m}}$ - the prime factorization of the number $n$. Any natural divisor of this number has the form $d=p_{1}^{h_{1}} \cdot p_{2}^{l_{2}} \cdot \ldots \cdot p_{m}^{l_{m}}$, where $l_{i} \in\left\{0,1, \ldots, k_{i}\right\}, i=1, \ldots, m$. The number of divisors of the number $n$ is $\left(k_{1}+1\right)\left(k_{2}+1\right) \cdots\left(k_{m}+1\right)=70$. We factorize the number 70 into non-unit factors in all possible ways and choose the smallest number $n$. Since $70=2 \cdot 5 \cdot 7$, we have five cases:

1) $70=70$, the smallest number $n=2^{69}>40000$;
2) $70=35 \cdot 2$, the smallest number $n=2^{34} \cdot 3^{1}>40000$;
3) $70=14 \cdot 5$, the smallest number $n=2^{13} \cdot 3^{4}>40000$;
4) $70=10 \cdot 7$, the smallest number $n=2^{9} \cdot 3^{6}=512 \cdot 81>40000$;
5) $70=7 \cdot 5 \cdot 2$, the smallest number $n=2^{6} \cdot 3^{4} \cdot 5^{1}=25920$.

Answer: 25920.