Determine all numbers congruent to 1 modulo 27 and to 6 modulo 37.
This is a direct application of the above demonstration. If $a_{1}=27$ and $a_{2}=37, N_{1}=37$ and $N_{2}=27$. To find the Bézout relation between 27 and 37, one possibility, as in Exercise 3, is to try all multiples of 27 until finding one congruent to 1 modulo 37, we find: $27 \times 11 - 37 \times 8 = 1$. Another possibility is to trace back the Euclidean algorithm: $37 = 27 + 10$, $27 = 2 \times 10 + 7$, $10 = 7 + 3$ and $7 = 2 \times 3 + 1$. By retracing:

$$
\begin{gathered}
1 = 7 - 2 \times 3 = 7 - 2 \times (10 - 7) = 3 \times 7 - 2 \times 10 = \\
= 3 \times (27 - 2 \times 10) - 2 \times 10 = 3 \times 27 - 8 \times 10 \\
= 3 \times 27 - 8 \times (37 - 27) = 11 \times 27 - 8 \times 37
\end{gathered}
$$

Once these $x_{i}$ are found, we simply apply the formula: $B = 1 \times (-8 \times 37) + 6 \times (11 \times 27) = 1486$. The numbers sought are all numbers congruent to $1486 \pmod{27 \times 37}$, or equivalently to $487 \pmod{999}$.