AMC 12B 2025 (Problem 3)What is the value of i(i−1)(i−2)(i−3)i(i-1)(i-2)(i-3)i(i−1)(i−2)(i−3), where i=−1i=\sqrt{-1}i=−1?(A) 6-5i\text{(A)}\;\text{6-5i}(A)6-5i(B) -10i\text{(B)}\;\text{-10i}(B)-10i(C) 10i\text{(C)}\;\text{10i}(C)10i(D) −10\text{(D)}\;-10(D)−10(E) 10\text{(E)}\;10(E)10Related TopicsCoreToolkit 97 — Basic Complex NumbersMajorToolkit 56 — Gauss' Idea (Rainbow Idea)Toolkit 10 — Difference of squaresHints (3)Hint 1T=i(i−1)(i−2)(i−3)=(i(i−3))((i−1)(i−2))T=i(i-1)(i-2)(i-3)=(i(i-3))((i-1)(i-2))T=i(i−1)(i−2)(i−3)=(i(i−3))((i−1)(i−2))Hint 2T=(i2−3i)(i2−3i+2)=(−1−3i)(−3i+1)T=(i^2-3i)(i^2-3i+2)=(-1-3i)(-3i+1)T=(i2−3i)(i2−3i+2)=(−1−3i)(−3i+1)Hint 3T=(−3i)2−1=9i2−1=−9−1=−10T=(-3i)^2-1=9i^2-1=-9-1=-10T=(−3i)2−1=9i2−1=−9−1=−10Final Answer(D) −10-10−10Related Problems (11)AMC 12A 2025 (Problem 9)AMC 12A 2025 (Problem 17)AMC 10A 2024 (Problem 11)AMC 10A/12A Spring 2021 (Problem 10/9)AMC 10A 2012 (Problem 22)AMC 10B 2023 (Problem 14)BMO1 2016/2017 (Problem 3)AMC 10A/12A 2024 (Problem 15/9)View all related problems →