2025 AMC 10B Problem 10

Subtract first; the cubic terms cancel:
$$f(n) - g(n) = n^2 - 3n - 4 = (n-4)(n+1).$$
That suggests testing $n = 4$ and $n = -1$ in $g$: $g(4) = 64 - 96 + 20 + 12 = 0$ and $g(-1) = -1 - 6 - 5 + 12 = 0$. So $(n-4)(n+1)$ divides $g$, and since $g$ is monic of degree $3$,
$$g(n) = (n-4)(n+1)(n-3).$$
Then $f(n) = g(n) + (n-4)(n+1) = (n-4)(n+1)(n-2)$.

For every $n$ outside $\{-1, 3, 4\}$ the common factor cancels:
$$\frac{f(n)}{g(n)} = \frac{n-2}{n-3} = 1 + \frac{1}{n-3}.$$
This is an integer exactly when $n - 3$ divides $1$, that is $n = 4$ or $n = 2$.

Now check the excluded values. At $n = -1, 3, 4$ the denominator $g(n)$ is zero, so the quotient does not exist; in particular $n = 4$ must be discarded even though it survived the cancelled form. Only $n = 2$ remains, and $\frac{f(2)}{g(2)} = \frac{0}{-6} = 0$ is indeed an integer.

The sum of all such $n$ is $2$.

The answer is $\boxed{\textbf{(A)}\ 2}$.

When a problem asks for integrality of a ratio of two polynomials of the same degree, the difference is the cheapest thing to compute: it has lower degree and its roots are the natural candidates for a common factor. Once the ratio is reduced to $\frac{n-2}{n-3}$, splitting off the integer part turns the question into a divisibility statement about a single small number. The one piece of bookkeeping the algebra hides is that cancelling $(n-4)(n+1)$ is legitimate only when those factors are non-zero, so the zeros of the original denominator must be checked separately.

Factor each cubic from scratch. Integer roots divide the constant term, so try the divisors of $8$ in $f$: $f(2) = 8 - 20 + 4 + 8 = 0$. Dividing gives $f(n) = (n-2)(n^2 - 3n - 4)$. Similarly $g(3) = 27 - 54 + 15 + 12 = 0$ gives $g(n) = (n-3)(n^2 - 3n - 4)$. The shared quadratic $n^2 - 3n - 4$ cancels, leaving the same $\frac{n-2}{n-3}$.

• Keeping n = 4 as a solution: the reduced fraction is an integer there, but g(4) = 0, so f(4)/g(4) is undefined; including it gives 6 at choice (E).
• Solving $n - 3 \mid 1$ as $n - 3 = 1$ only and forgetting $n - 3 = -1$, or vice versa.
• Treating $\frac{0}{-6}$ at $n = 2$ as undefined and discarding the one value that does work.