6. Let $n$ be the answer to this problem. $a$ and $b$ are positive integers satisfying
$$
\begin{array}{l}
3 a+5 b \equiv 19 \quad(\bmod n+1) \\
4 a+2 b \equiv 25 \quad(\bmod n+1)
\end{array}
$$

Find $2 a+6 b$.
Proposed by: David Vulakh
Answer: 96
Solution: Let $m=n+1$, so that the conditions become
$$
\begin{array}{l}
3 a+5 b \equiv 19 \quad(\bmod m) \\
4 a+2 b \equiv 25 \quad(\bmod m) \\
2 a+6 b \equiv-1 \quad(\bmod m)
\end{array}
$$

We can subtract (2) from twice (3) to obtain
$$
10 b \equiv-27 \quad(\bmod m)
$$

Multiplying (1) by 2 and replacing $10 b$ with -27 gives
$$
6 a-27 \equiv 38 \quad(\bmod m)
$$

So $6 a \equiv 65(\bmod m)$. Multiplying (3) by 30 and replacing $10 b$ with -27 and $6 a$ with 64 gives
$$
650-486 \equiv-30 \quad(\bmod m)
$$

Therefore $194 \equiv 0(\bmod m)$, so $m \mid 194$. Since the prime factorization of $m$ is $194=97 \cdot 2, m$ must be 1 , 2,97 , or 194. Condition (3) guarantees that $m$ is odd, and $a, b>0$ guarantees that $m=2 a+6 b+1 \neq 1$.

So we must have $m=97$, so $n=96$. A valid solution is $a=27, b=7$.