HMMT February, 2024

Solution:
Let $f(x) = \frac{p(x)}{q(x)}$. Then, $x^{3} q(x) - p(x)$ has $1,2,3,4,5$ as roots. Therefore, WLOG, let

$$x^{3} q(x) - p(x) = (x-1)(x-2)(x-3)(x-4)(x-5) = x^{5} - 15 x^{4} + 85 x^{3} - \ldots$$

Thus, $q(x) = x^{2} - 15 x + 85$, so $q(0) = 85$. Plugging $x = 0$ in the above equation also gives $-p(0) = -120$. Hence, the answer is $\frac{120}{85} = \frac{24}{17}$.

Remark. From the solution above, it is not hard to see that the unique $f$ that satisfies the problem is
$$f(x) = \frac{225 x^{2} - 274 x + 120}{x^{2} - 15 x + 85}$$