AMC 10A/12A Spring 2021 (Problem 14/12)All the roots of the polynomial z6−10z5+Az4+Bz3+Cz2+Dz+16z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16z6−10z5+Az4+Bz3+Cz2+Dz+16 are positive integers, possibly repeated. What is the value of BBB?(A) −88\text{(A)}\;-88(A)−88(B) −80\text{(B)}\;-80(B)−80(C) −64\text{(C)}\;-64(C)−64(D) −41\text{(D)}\;-41(D)−41(E) −40\text{(E)}\;-40(E)−40Related TopicsCoreVieta's FormulaCheck AnswerYour answer:ABCDECheckHints (4)Hint 1Assume r1,r2,…,r6r_1,r_2,\ldots,r_6r1,r2,…,r6 are the roots of z6−10z5+Az4+Bz3+Cz2+Dz+16z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16z6−10z5+Az4+Bz3+Cz2+Dz+16.By Vieta's Formula,r1+r2+r3+r4+r5+r6=10r_1+r_2+r_3+r_4+r_5+r_6=10r1+r2+r3+r4+r5+r6=10r1r2r3r4r5r6=16r_1r_2r_3r_4r_5r_6=16r1r2r3r4r5r6=16Hint 2The only possibility with positive integer roots is2,2,2,2,1,1.2,2,2,2,1,1.2,2,2,2,1,1.Hint 3By Vieta's Formula,r1r2r3+r1r2r4+⋯+r4r5r6=−Br_1r_2r_3+r_1r_2r_4+\cdots+r_4r_5r_6=-Br1r2r3+r1r2r4+⋯+r4r5r6=−BHint 4The roots are 1,1,2,2,2,21,1,2,2,2,21,1,2,2,2,2.B=−(r1r2r3+r1r2r4+⋯+r4r5r6)B=-\left(r_1r_2r_3+r_1r_2r_4+\cdots+r_4r_5r_6\right)B=−(r1r2r3+r1r2r4+⋯+r4r5r6)=−[(43)(2⋅2⋅2)+(42)(21)(2⋅2⋅1)+(41)(22)(2⋅1⋅1)]=-\left[\binom43(2\cdot2\cdot2)+\binom42\binom21(2\cdot2\cdot1)+\binom41\binom22(2\cdot1\cdot1)\right]=−[(34)(2⋅2⋅2)+(24)(12)(2⋅2⋅1)+(14)(22)(2⋅1⋅1)]=−(32+48+8)=−88=-(32+48+8)=-88=−(32+48+8)=−88Final Answer(A) −88-88−88Related Problems (10)AMC 10A/12A 2022 (Problem 16/15)AMC 12B 2023 (Problem 14)AMC 12B 2022 (Problem 4)AMC 12B 2021 Spring (Problem 16)AMC 10A/12A 2025 (Problem 18/12)AMC 12A 2025 (Problem 19)AMC 12A 2024 (Problem 15)AMC 12B 2024 (Problem 17)AIME II 2026 (Problem 11)AMC 12B 2025 (Problem 8)View all related problems →