Example 8 Find a 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 $N$ in its prime factorization form
$N=2^{a_{1}} \cdot 3^{a_{2}} \cdot 5^{a_{3}} \cdot 7^{a_{4} \cdots} p_{n n}^{a_{n}}$, where $a_{i} \geqslant 0, i=1,2, \cdots, n$, then the number of all divisors of it is $\left(a_{1}+1\right)\left(a_{2}+1\right)\left(a_{3}+1\right)\left(a_{1}+1\right) \cdots\left(a_{n}+1\right)=10$.
Since $5\left|N, 7^{2}\right| N$, then $a_{3}+1 \geqslant 2, a_{1}+1 \geqslant 3$,
Therefore, $a_{1}, a_{2}, a_{5}, a_{6}, \cdots, a_{n}$ must all be 0.
Thus, $N=5^{a_{3}} \cdot 7^{a_{4}}$.
Since $\left(a_{3}+1\right)\left(a_{4}+1\right)=10=2 \times 5$, we get $a_{3}=1, a_{4}=4$, so the unique solution to this problem is $N=5 \cdot 7^{4}$.