2025 AMC 10B Problem 10

P1-10~4 minPrint

Polynomials Divisibility & Factors Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Set up the equation/formula and compute; no special trick needed

Let $f(n)=n^3-5n^2+2n+8$ and $g(n)=n^3-6n^2+5n+12.$ What is the sum of all integers $n$ such that $\tfrac{f(n)}{g(n)}$ is an integer?

Show solution without answering

Key insight

Subtracting gives f − g = (n−4)(n+1), a factor of both. The fraction becomes (n−2)/(n−3), but n = 4 is out because g(4) = 0.

Solution

1. Subtract to find something in common. The two cubics have the same $n^3$ term, so subtract them and it disappears:
$f(n) - g(n) = (n^3 - 5n^2 + 2n + 8) - (n^3 - 6n^2 + 5n + 12) = n^2 - 3n - 4 = (n-4)(n+1).$

2. Test those numbers in $g$ . Try $n = 4$ : $g(4) = 64 - 96 + 20 + 12 = 0$ . Try $n = -1$ : $g(-1) = -1 - 6 - 5 + 12 = 0$ . So $(n-4)$ and $(n+1)$ are both factors of $g$ . The third factor must make the constant term $12$ : $(-4)(1)(\,?\,) = 12$ , so it is $(n - 3)$ :
$g(n) = (n-4)(n+1)(n-3).$

3. Factor $f$ the same way. $f = g + (n-4)(n+1)$ , so
$f(n) = (n-4)(n+1)\bigl[(n-3) + 1\bigr] = (n-4)(n+1)(n-2).$

4. Simplify the fraction. When $n$ is not $4$ , $-1$ or $3$ , cancel the common factors:
$\frac{f(n)}{g(n)} = \frac{n-2}{n-3} = \frac{(n-3) + 1}{n-3} = 1 + \frac{1}{n-3}.$
This is a whole number only when $\frac{1}{n-3}$ is, which means $n - 3 = 1$ or $n - 3 = -1$ . So $n = 4$ or $n = 2$ .

Check the numbers where we cancelled. At $n = 4$ , $-1$ and $3$ , the original denominator $g(n)$ is $0$ , so the fraction does not exist at all. That rules out $n = 4$ , even though the simplified fraction looked fine there.

Check $n = 2$ . $f(2) = 8 - 20 + 4 + 8 = 0$ and $g(2) = 8 - 24 + 10 + 12 = 6$ , so the fraction is $0$ , a whole number.

Only $n = 2$ works, so the sum is $2$ .

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

Why this works

When two polynomials start the same way, subtracting them gives something smaller that often shows their shared factors. After cancelling, pulling out the whole-number part ( $1 + \frac{1}{n-3}$ ) turns "is this a whole number?" into "what divides $1$ ?". Whenever you cancel a factor, go back and check the values that made it zero; the original fraction may not exist there.

Alternative approach

Look for whole-number roots of each cubic. A whole-number root must divide the constant term. For $f$ , try divisors of $8$ : $f(2) = 0$ , and dividing $f$ by $(n - 2)$ leaves $n^2 - 3n - 4$ . For $g$ , try divisors of $12$ : $g(3) = 27 - 54 + 15 + 12 = 0$ , and dividing $g$ by $(n-3)$ leaves the same $n^2 - 3n - 4$ . Cancel it to get $\frac{n-2}{n-3}$ , then finish as above.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(A)}\ 2}$ .

Review in the AoPS books

Intermediate Algebra, Chapter 6: Polynomial Division (§6.2, §6.4)
Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Algebra, Chapter 11: Special Factorizations
Intermediate Algebra, Chapter 7: Polynomial Roots Part I (§7.1, §7.2, §7.3)