2002 AMC 10B Problem 1Problem 1 of 25·Easier·ArithmeticWhat is the value of the ratio 22001⋅3200362002?\dfrac{2^{2001} \cdot 3^{2003}}{6^{2002}}?6200222001⋅32003?Answer choicesA16\dfrac{1}{6}611B13\dfrac{1}{3}312C12\dfrac{1}{2}213D23\dfrac{2}{3}324E32\dfrac{3}{2}235Submit answerStuck? Show hintsShow solutionSolutionSince 62002=22002⋅32002,6^{2002} = 2^{2002}\cdot 3^{2002},62002=22002⋅32002, the ratio becomes 22001⋅3200322002⋅32002=32.\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}} = \dfrac{3}{2}.22002⋅3200222001⋅32003=23. Thus, the correct answer is E.AoPS wikiCopy problemTagged: exponent