$4 \cdot 6$ Find a natural number $N$, such that it is divisible by 5 and 49, and including 1 and $N$ itself, it has exactly 10 divisors.
(Kyiv Mathematical Olympiad, 1982)
[Solution] Write the number $N$ in its prime factorization form
$$
N=2^{\alpha_{1}} \cdot 3^{\alpha_{2}} \cdot 5^{\alpha_{3}} \cdot 7^{\alpha_{4} \cdots} p_{n}^{\alpha_{n}} .
$$

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

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

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

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

we get
$$
\alpha_{3}=1, \quad \alpha_{4}=4 \text {, }
$$

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