AMC 10B 2025 (Problem 10)Let f(n)=n3−5n2+2n+8f(n)=n^3-5n^2+2n+8f(n)=n3−5n2+2n+8 and g(n)=n3−6n2+5n+12g(n)=n^3-6n^2+5n+12g(n)=n3−6n2+5n+12. What is the sum of all integers nnn such that f(n)g(n)\frac{f(n)}{g(n)}g(n)f(n) is an integer?(A) 2\text{(A)}\;2(A)2(B) 3\text{(B)}\;3(B)3(C) 4\text{(C)}\;4(C)4(D) 5\text{(D)}\;5(D)5(E) 6\text{(E)}\;6(E)6Related TopicsCoreToolkit 100 — Factor TheoremMajorToolkit 19 — Factoring quadratic by groupingMinorToolkit 46 — Divisibility: Definition and PropertiesHints (9)Hint 1Factorize f(n),g(n)f(n),g(n)f(n),g(n).Hint 2f(−1)=(−1)3−5(−1)2+2(−1)+8=−1−5−2+8=0f(-1)=(-1)^3-5(-1)^2+2(-1)+8=-1-5-2+8=0f(−1)=(−1)3−5(−1)2+2(−1)+8=−1−5−2+8=0By Toolkit 100 — Factor Theorem, n+1n+1n+1 is a factor of f(n)f(n)f(n).Hint 3f(n)=n3−5n2+2n+8=(n+1)(n2−6n+8)f(n)=n^3-5n^2+2n+8=(n+1)(n^2-6n+8)f(n)=n3−5n2+2n+8=(n+1)(n2−6n+8)Hint 4f(n)=(n+1)(n2−6n+8)=(n+1)(n−2)(n−4)f(n)=(n+1)(n^2-6n+8)=(n+1)(n-2)(n-4)f(n)=(n+1)(n2−6n+8)=(n+1)(n−2)(n−4)Hint 5g(−1)=(−1)3−6(−1)2+5(−1)+12=−1−6−5+12=0g(-1)=(-1)^3-6(-1)^2+5(-1)+12=-1-6-5+12=0g(−1)=(−1)3−6(−1)2+5(−1)+12=−1−6−5+12=0By Toolkit 100 — Factor Theorem, n+1n+1n+1 is a factor of g(n)g(n)g(n).Hint 6g(n)=n3−6n2+5n+12=(n+1)(n2−7n+12)g(n)=n^3-6n^2+5n+12=(n+1)(n^2-7n+12)g(n)=n3−6n2+5n+12=(n+1)(n2−7n+12)Hint 7g(n)=(n+1)(n2−7n+12)=(n+1)(n−3)(n−4)g(n)=(n+1)(n^2-7n+12)=(n+1)(n-3)(n-4)g(n)=(n+1)(n2−7n+12)=(n+1)(n−3)(n−4)Hint 8By Hints 4 and 7:f(n)g(n)=(n+1)(n−2)(n−4)(n+1)(n−3)(n−4)=n−2n−3,n≠−1,3,4\frac{f(n)}{g(n)}=\frac{(n+1)(n-2)(n-4)}{(n+1)(n-3)(n-4)}=\frac{n-2}{n-3},\qquad n\ne-1,3,4g(n)f(n)=(n+1)(n−3)(n−4)(n+1)(n−2)(n−4)=n−3n−2,n=−1,3,4Hint 9f(n)g(n) is an integer\frac{f(n)}{g(n)}\text{ is an integer}g(n)f(n) is an integer⇒n−3∣n−2\Rightarrow n-3\mid n-2⇒n−3∣n−2n−3∣n−3n-3\mid n-3n−3∣n−3⇒n−3∣1⇒n−3=±1\Rightarrow n-3\mid1\Rightarrow n-3=\pm1⇒n−3∣1⇒n−3=±1⇒n=3±1=2,4\Rightarrow n=3\pm1=2,4⇒n=3±1=2,4By Hint 8, n≠4⇒n=2n\ne4\Rightarrow n=2n=4⇒n=2Final Answer(A) 222Related Problems (3)AMC 10A 2020 (Problem 5)AMC 12A 2025 (Problem 17)AMC 10B 2025 (Problem 15)