3. For which numbers $n$ are the numbers $a_{n}=7 n-3$ divisible by 5 but not divisible by 3?
Solution. $a_{n}=7 n-3: 5 \rightarrow 7 n-3=5 k, k \in Z$. We obtain the linear Diophantine equation $7 n-5 k=3, k \in Z$. We have

$$
\begin{aligned}
& k=\frac{7 n-3}{5}=n+\frac{2 n-3}{5} \rightarrow \frac{2 n-3}{5}=l \in Z \rightarrow n=\frac{5 l+3}{2}=2 l+1+\frac{l+1}{2} \rightarrow \\
& \rightarrow \frac{l+1}{2}=t \in Z \rightarrow l=2 t-1 \rightarrow n=2(2 t-1)+t=5 t-1 .
\end{aligned}
$$

We obtained $n=5 t-1, t \in N$. Then, $a_{n}=35 t-10, t \in N$. Rewrite $a_{n}$ as $a_{n}=36 t-9-t-1, t \in N$. Therefore, $a_{n} \vdots 3 \leftrightarrow t+1: 3 \leftrightarrow t=3 m-1, m \in N$. The problem's condition is satisfied by the numbers $n=5 t-1, t \neq 3 m-1, t, m \in N$.

Answer: $n=5 t-1, t \neq 3 m-1, t, m \in N$.