Track / Stage 4 / 274 of 340 #534 of 1964

Problem 534

AMC 12 late, AIME early

Algebra Difficulty 4.9 Multiple choice

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

Pick one

Official solution

［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 $(D)$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.