19. [11] Integers $0 \leq a, b, c, d \leq 9$ satisfy
$$
\begin{array}{c}
6 a+9 b+3 c+d=88 \\
a-b+c-d=-6 \\
a-9 b+3 c-d=-46
\end{array}
$$

Find $1000 a+100 b+10 c+d$.
Answer: 6507
Solution: Let $N=\overline{a b c d}$ be an at most-four digit number. Note that the first equation gives us $N \equiv 4(\bmod 7)$. The second equation gives us $N \equiv 6(\bmod 11)$. The third equation gives us $N \equiv 7$ $(\bmod 13)$. Using CRT, we get $N \equiv \frac{1}{2} \equiv 501(\bmod 1001)$. Thus, we have $N=501+1001 k$ for some integer $0 \leq k \leq 9$. The only value of $k$ that satisfies the first equation is $k=6$, which yields $N=6507$.