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2021 AMC 10A Problem 10

Problem 10 of 25EasierAlgebra

Which of the following is equivalent to (2+3)(22+32)(24+34)(28+38)(216+316)(232+332)(264+364)? \begin{aligned} &(2+3)(2^2+3^2)\\ &\quad\cdot(2^4+3^4)(2^8+3^8)\\ &\quad\cdot(2^{16}+3^{16})(2^{32}+3^{32})\\ &\quad\cdot(2^{64}+3^{64})? \end{aligned}

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Solution

Multiply the product by 32=1.3-2=1. Repeatedly applying the difference-of-squares identity gives (32)(3+2)=3222,(3222)(32+22)=3424, \begin{aligned} (3-2)(3+2)&=3^2-2^2,\\ (3^2-2^2)(3^2+2^2)&=3^4-2^4, \end{aligned} and the same cancellation continues through the final factor. Therefore the product is 31282128.3^{128}-2^{128}. Thus, C is the correct answer.

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Concepts: difference of squares · telescoping

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.