295. Find all natural numbers $n$ for which the number $7 n+5$ is divisible by the number $3 n-1$.
$\triangle$ Here, the straightforward division of the dividend by the divisor using the "corner" method does not work, as the number $7 / 3$ is not an integer. Let's try an alternative approach.

If the fraction $\frac{7 n+5}{3 n-1}$ is a natural number, then the tripled fraction

$$
\frac{3(7 n+5)}{3 n-1}=\frac{21 n+15}{3 n-1}
$$

is also a natural number. The converse statement is also true, since the numbers 3 and $3 n-1$ are coprime (think about why).

Now we can apply either the "corner" division method or the "plus-minus" method. In the latter case, we need to know which term to isolate in the numerator of the last fraction. Clearly, the term is

$$
(3 n-1) \cdot 7=21 n-7
$$

We get:

$$
\frac{21 n+15}{3 n-1}=\frac{(21 n-7)+22}{3 n-1}=7+\frac{22}{3 n-1}
$$

It remains to check the cases when $3 n-1$ equals $1, 2, 11$, and 22.

Answer: $1, 4$.