9.5. The sum of some consecutive odd natural numbers is 392. Find these numbers.
Solution. Let these numbers be: $2 m+1,2 m+3,2 m+5, \ldots . ., 2 m+1+2 k$ We find the numbers $m, k$ from the relationship

$$
2 m+1+2 m+3+2 m+5+\cdots+2 m+1+2 k=392
$$

or

$$
\frac{2 m+1+2 m+1+2 k}{2} \cdot(k+1)=392 \Leftrightarrow(2 m+k+1)(k+1)=392=2^{3} \cdot 7^{2}
$$

If $k+1$ were odd, then $2 m+k+1$ would also be odd, and thus their product would be odd, which contradicts the equation $(2 m+k+1)(k+1)=392$. Therefore, $k+1$ is even, and so is $2 m+k+1$. Considering that $392=2^{3} \cdot 7^{2}, \quad(2 m+k+1)(k+1)=2^{3} \cdot 7^{2}$ and that $2 m+k+1>k+1$, we obtain only the following 3 cases:

| $2 m+k+1$ | $k+1$ | $m$ | $k$ | The numbers obtained |
| :--- | :--- | :--- | :--- | :--- |
| $196=2^{2} \cdot 7^{2}$ | 2 | 97 | 1 | 195,197 |
| $98=2 \cdot 7^{2}$ | 4 | 47 | 3 | $95,97,99,101$ |
| $28=2^{2} \cdot 7$ | 14 | 7 | 13 | $15,17,19,21,23,25,27,29,31,33,35,37,39,41$ |