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Problem 570

Algebra Difficulty 2.6 Multiple choice CEMC Fermat · Canada · 2012

The expression 32011+3201132010+32012\dfrac{3^{2011}+3^{2011}}{3^{2010}+3^{2012}} is equal to

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Official solution

Since 32011=31⋅32010=3⋅320103^{2011} = 3^1 \cdot 3^{2010} = 3\cdot 3^{2010} and 32012=32⋅32010=9⋅320103^{2012} = 3^2 \cdot 3^{2010} = 9\cdot3^{2010}, then 32011+3201132010+32012=3⋅32010+3⋅3201032010+9⋅32010=32010(3+3)32010(1+9)=3+31+9=610=35\dfrac{3^{2011} + 3^{2011}}{3^{2010}+3^{2012}} = \dfrac{3\cdot 3^{2010} + 3\cdot 3^{2010}}{3^{2010} + 9\cdot 3^{2010}} = \dfrac{3^{2010}(3+3)}{3^{2010}(1+9)} = \dfrac{3+3}{1+9} = \dfrac{6}{10} = \dfrac{3}{5}

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.