$12 \cdot 81$ Given $x=-2272, y=10^{3}+10^{2} c+10 b+a, z=1$ satisfies the equation $a x+b y+c z=1$, where $a, b, c$ are positive integers, $a<b<c$. Find $y$.
(28th International Mathematical Olympiad Candidate Question, 1987)
[Solution] From the given,
$$
b(1000+100 c+10 b+a)+c-2272 a-1=0 .
$$

If $b \geqslant a+2$, then $c \geqslant a+3$, thus
$$
\begin{aligned}
0 & \geqslant(a+2)[1000+100(a+3)+10(a+2)+a] \\
& +(a+3)-2272 a-1 \\
& =111 a^{2}-729 a+2642 .
\end{aligned}
$$

Let $u=111 a^{2}-729 a+2642$, then
$$
u \geqslant \frac{4 \cdot 111 \cdot 2642-729^{2}}{4 \cdot 111}>0 \text {. }
$$

This leads to a contradiction $0>0$.
Therefore, $b=a+1$.
Let $c=a+2+t, t \geqslant 0$, then
$$
\begin{aligned}
0= & (a+1)[1000+100(a+2+t)+10(a+1)+a] \\
& +a+2+t-2272 a-1,
\end{aligned}
$$

which simplifies to $111 a^{2}+(100 t-950) a+(1211+101 t)=0$.
If $t \geqslant 2$, then
$$
\begin{aligned}
0 & \geqslant 111 a^{2}+(200-950) a+(1211+202) \\
& =111 a^{2}-750 a+1413 .
\end{aligned}
$$

Let $v=111 a^{2}-750 a+1413$, then
$$
v \geqslant \frac{4 \cdot 111 \cdot 1413-750^{2}}{4 \cdot 111}>0 \text {. }
$$

This again leads to a contradiction $0>0$.
Therefore, $t=0$ or 1.
When $t=1$, $b=a+1, c=a+3$, then
$$
111 a^{2}-850 a+1312=0 .
$$

This equation has no integer solutions.
When $t=0$, $b=a+1, c=a+2$, then
$$
\begin{array}{l}
111 a^{2}-950 a+1211=0, \\
(a-7)(111 a-173)=0,
\end{array}
$$

Thus,
$$
a=7, \quad b=8, \quad c=9 .
$$

Therefore,
$$
y=1987
$$