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Problem 405 Difficulty 2.6 Multiple choice CEMC Fermat · Canada · 2012
The expression 3 2011 + 3 2011 3 2010 + 3 2012 \dfrac{3^{2011}+3^{2011}}{3^{2010}+3^{2012}} 3 2010 + 3 2012 3 2011 + 3 2011
Pick one
A 3 5 \frac{3}{5} 5 3 B 1 1 1 C 9 10 \frac{9}{10} 10 9 D 10 3 \frac{10}{3} 3 10 E 2 3 \frac{2}{3} 3 2
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Official solution Since 3 2011 = 3 1 ⋅ 3 2010 = 3 ⋅ 3 2010 3^{2011} = 3^1 \cdot 3^{2010} = 3\cdot 3^{2010} 3 2011 = 3 1 ⋅ 3 2010 = 3 ⋅ 3 2010 3 2012 = 3 2 ⋅ 3 2010 = 9 ⋅ 3 2010 3^{2012} = 3^2 \cdot 3^{2010} = 9\cdot3^{2010} 3 2012 = 3 2 ⋅ 3 2010 = 9 ⋅ 3 2010 3 2011 + 3 2011 3 2010 + 3 2012 = 3 ⋅ 3 2010 + 3 ⋅ 3 2010 3 2010 + 9 ⋅ 3 2010 = 3 2010 ( 3 + 3 ) 3 2010 ( 1 + 9 ) = 3 + 3 1 + 9 = 6 10 = 3 5 \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} 3 2010 + 3 2012 3 2011 + 3 2011 = 3 2010 + 9 ⋅ 3 2010 3 ⋅ 3 2010 + 3 ⋅ 3 2010 = 3 2010 ( 1 + 9 ) 3 2010 ( 3 + 3 ) = 1 + 9 3 + 3 = 10 6 = 5 3
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