$31 \cdot 18$ A child, whose age is between 13 and 19, writes his own age after his father's age. From this new four-digit number, subtract the absolute value of their age difference to get 4289. The sum of their ages is 
(A) 48.
(B) 52.
(C) 56.
(D) 59.
(E) 64.
(22nd American High School Mathematics Examination, 1971)
[Solution] Let $b$ and $f$ represent the ages of the child and the father, respectively. Then $13 \leqslant b \leqslant 19$, and $100 f + b - (f - b) = 99 f + 2 b = 4289$. That is, $99 f = (99 \cdot 43 + 32) - 2 b$, or $f - 43 = \frac{32 - 2 b}{99}$.
If $f = 43$, then $32 = 2 b$, which means $b = 16$;
If $f \geqslant 44$, then $b > 65$, which is not reasonable.
In summary, $f = 43, b = 16$, and $f + b = 59$.
Therefore, the answer is $(D)$.