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2002 AMC 10B Problem 1

Problem 1 of 25EasierArithmetic

What is the value of the ratio 22001⋅3200362002?\dfrac{2^{2001} \cdot 3^{2003}}{6^{2002}}?

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Solution

Since 62002=22002⋅32002,6^{2002} = 2^{2002}\cdot 3^{2002}, the ratio becomes 22001⋅3200322002⋅32002=32.\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}} = \dfrac{3}{2}. Thus, the correct answer is E.
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