2025 AMC 10B Problem 10

Question

Let

f(n) = n^3 - 5n^2 + 2n + 8,

and let

g(n) = n^3 - 6n^2 + 5n + 12.

What is the sum of all integer values of

n

for which

\dfrac{f(n)}{g(n)}

is also an integer?

2

3

4

5

6

Both cubics have

n = -1

as a root; factor each and cancel common factors

The ratio simplifies to

\dfrac{n - 2}{n - 3} = 1 + \dfrac{1}{n - 3};

it is an integer only when

n - 3

divides

1

Factor both cubics:

f(n) = (n + 1)(n - 2)(n - 4)

and

g(n) = (n + 1)(n - 3)(n - 4).

Away from

n \in \{-1, 3, 4\},

where

g

vanishes or the ratio is

\tfrac{0}{0},

the common factors cancel and

\dfrac{f(n)}{g(n)} = \dfrac{n - 2}{n - 3} = 1 + \dfrac{1}{n - 3}.

That’s an integer only when

n - 3 = \pm 1,

so

n = 2

or

n = 4.

But

n = 4

kills

g,

so only

n = 2

survives, and the sum is

2.

Therefore, the answer is A.

Problem 10 in other years