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2021 AMC 12A Problem 9

Problem 9 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)(2^4 + 3^4) \\ &\quad {}\cdot (2^8 + 3^8)(2^{16} + 3^{16}) \\ &\quad {}\cdot (2^{32} + 3^{32})(2^{64} + 3^{64})? \end{aligned}

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Solution

Since 32=1,3 - 2 = 1, multiplying the product by 323 - 2 does not change it. Then (32)(3+2)=3222, (3-2)(3+2) = 3^2 - 2^2, and multiplying by the next factor (32+22)(3^2 + 2^2) gives 3424,3^4 - 2^4, and so on. Each step doubles the exponent. After using all seven factors, the product telescopes to 31282128.3^{128} - 2^{128}. Thus, the correct answer is C.

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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.