AMC 10/12A Spring 2021 (Problem 10)Which of the following is equivalent to (2+3)(22+32)(24+34)(28+38)(216+316)(232+332)(264+364)(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})(2+3)(22+32)(24+34)(28+38)(216+316)(232+332)(264+364)?(A) 3127+2127\text{(A)}\;3^{127}+2^{127}(A)3127+2127(B) 3127+2127+2⋅363+3⋅263\text{(B)}\;3^{127}+2^{127}+2\cdot3^{63}+3\cdot2^{63}(B)3127+2127+2⋅363+3⋅263(C) 3128−2128\text{(C)}\;3^{128}-2^{128}(C)3128−2128(D) 3128+2128\text{(D)}\;3^{128}+2^{128}(D)3128+2128(E) 5127\text{(E)}\;5^{127}(E)5127Related TopicsCoreToolkit 10 — Difference of squaresHints (2)Hint 1Multiply by (3−2)(3-2)(3−2) and use Toolkit 10 — Difference of squares several times.Hint 2Since 3−2=13-2=13−2=1, multiplying by (3−2)(3-2)(3−2) has no effect.(3−2)(2+3)(22+32)(24+34)⋯(264+364)(3-2)(2+3)(2^2+3^2)(2^4+3^4)\cdots(2^{64}+3^{64})(3−2)(2+3)(22+32)(24+34)⋯(264+364)=(32−22)(22+32)(24+34)⋯(264+364)=(3^2-2^2)(2^2+3^2)(2^4+3^4)\cdots(2^{64}+3^{64})=(32−22)(22+32)(24+34)⋯(264+364)=(34−24)(24+34)⋯(264+364)=(3^4-2^4)(2^4+3^4)\cdots(2^{64}+3^{64})=(34−24)(24+34)⋯(264+364)=(38−28)(28+38)⋯(264+364)=(3^8-2^8)(2^8+3^8)\cdots(2^{64}+3^{64})=(38−28)(28+38)⋯(264+364)⋮\qquad\vdots⋮=3128−2128=3^{128}-2^{128}=3128−2128Final Answer(C) 3128−21283^{128}-2^{128}3128−2128Related Problems (6)AMC 10A 2024 (Problem 11)AMC 10A 2012 (Problem 22)AMC 10B 2023 (Problem 14)BMO1 2016/2017 (Problem 3)AMC 10A/12A 2024 (Problem 15/9)AMC 10A 2023 (Problem 23)