Example 7 (1982 Kyiv Mathematical Olympiad) Find the natural number $N$, such that it is divisible by 5 and 49, and including 1 and $N$, it has a total of 10 divisors.
Solve: Write the number $N$ in the form of prime factorization:
$$
N=2^{a_{1}} \cdot 3^{a_{2}} \cdot 5^{a_{3}} \cdot 7^{a_{4}} \cdots \cdots \cdot p_{n}^{a_{n}},
$$

where $\alpha_{i} \geqslant 0, i=1,2, \cdots, n$.
Then the number of all divisors of it is
$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right) \cdot \cdots \cdot\left(\alpha_{n}+1\right)=10 \text {. }
$$

Since $5\left|N, 7^{2}\right| N$, then
$$
\alpha_{3}+1 \geqslant 2, \alpha_{4}+1 \geqslant 3 \text {. }
$$

Therefore, $\alpha_{1}, \alpha_{2}, \alpha_{3}, \cdots, \alpha_{n}$ must all be equal to 0, i.e.,
$$
N=5^{\alpha_{3}} \cdot 7^{\alpha_{4}} \text {. }
$$

By $\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right)=10=2 \cdot 5$, we get
$$
\alpha_{3}=1, \alpha_{4}=4 \text {. }
$$

Thus, the unique solution to this problem is $N=5 \cdot 7^{4}$.