$31 \cdot 51$ satisfies the following equations:
$$
|a+b|+c=19 \text { and } a b+|c|=97
$$

The number of ordered integer triples $(a, b, c)$ is
(A) 0.
(B) 4.
(C) 6.
(D) 10.
(E) 12.
(48th American High School Mathematics Examination, 1997)
[Solution] (1) When $c \geqslant 0$, $|c|=c$, then from the given two equations we have
$$
|a+b|+78-a b=0 \text {. }
$$
i) If $a+b \geqslant 0$, equation $(*)$ becomes $(a-1)(b-1)=79$.
We get $\left\{\begin{array}{l}a=2, \\ b=80\end{array}\right.$ or $\left\{\begin{array}{l}a=80, \\ b=2 .\end{array}\right.$ Substituting these into the given equations contradicts $c \geqslant 0$.
ii) If $a+b<0$, equation $(*)$ becomes $(a+1)(b+1)=79$.
We get $\left\{\begin{array}{l}a=0, \\ b=78\end{array}\right.$ or $\left\{\begin{array}{l}a=78, \\ b=0 .\end{array}\right.$ Substituting these into the given equations contradicts $c \geqslant 0$.
(2) When $c<0$, $|c|=-c$, then from the given two equations we have
$$
|a+b|+a b-116=0 \quad \text { (**) }
$$
i) If $a+b \geqslant 0$, equation $(**)$ becomes $(a+1)(b+1)=117=9 \times 13=3 \times 39=1 \times 117$. We get
$(a, b)=(8,12)$ or $(12,8)$ or $(2,38)$ or $(38,2)$ or $(0,116)$ or $(116,0)$.
Substituting these into the given equations, we obtain
$(a, b, c)=(8,12,-1)$ or $(12,8,-1)$ or $(2,38,-21)$ or $(38,2,-21)$ or $(0,116,-97)$ or $(116,0,-97)$.
ii) If $a+b<0$, then equation $(**)$ becomes $(a-1)(b-1)=117=(-9) \times (-13)=(-3) \times (-39)=(-1) \times (-117)$. We get
$(a, b)=(-8,-12)$ or $(-12,-8)$ or $(-2,-38)$ or $(-38,-2)$ or $(0,-116)$ or $(-116,0)$
Substituting these into the given equations, we obtain
$(a, b, c)=(-8,-12,-1)$ or $(-12,-8,-1)$ or $(-2,-38,-21)$ or $(-38,-2,-21)$ or $(0,-116,-97)$ or $(-116,0,-97)$.
Combining (1) and (2), we find a total of 12 sets of $(a, b, c)$ that satisfy the given conditions. Therefore, the answer is $(E)$.