1.119 $A$ algebra books (all algebra books have the same thickness) and $H$ geometry books (all geometry books have the same thickness and are thicker than algebra books) exactly fill a bookshelf. It is also known that $S$ algebra books and $M$ geometry books can fill the same bookshelf. Finally, placing $E$ algebra books alone can also fill the same bookshelf. Given that $A, H, S, M, E$ are different positive integers, then $E$ equals
(A) $\frac{A M+S H}{M+H}$.
(B) $\frac{A M^{2}+S H^{2}}{M^{2}+H^{2}}$.
(C) $\frac{A H-\frac{S M}{H}}{M-}$.
(D) $\frac{A M-S H}{M-H}$.
(E) $\frac{A M^{2}-S H^{2}}{M^{2}-H^{2}}$.
(38th American High School Mathematics Examination, 1987)
［Solution］Let the thickness of the algebra book be $d$, and the thickness of the geometry book be $e$. According to the problem, we have
$$
A d+H e=S d+M e=E d,
$$

Solving this, we get $e=\frac{E-A}{H} d$, and $e=\frac{E-S}{M} d$.
$$
\therefore \quad \frac{E-A}{H}=\frac{E-S}{M}, \text{ i.e., } E=\frac{A M-S H}{M-H} \text{. }
$$

Therefore, the answer is $(D)$.