2002 AMC 10B Problems/Problem 1: Difference between revisions
New page: == Problem == The ratio <math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}</math> is: <math> \mathrm{(A) \ } 1/6\qquad \mathrm{(B) \ } 1/3\qquad \mathrm{(C) \ } 1/2\qquad \mathrm{(D) \ } 2/3\q... |
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<math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}=\frac{6^{2001}\cdot 3^2}{6^{2002}}=\frac{9}{6}=\frac{3}{2}</math> or <math>\mathrm{ (E) \ }</math> | <math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}=\frac{6^{2001}\cdot 3^2}{6^{2002}}=\frac{9}{6}=\frac{3}{2}</math> or <math>\mathrm{ (E) \ }</math> | ||
==See Also== | |||
{{AMC10 box|year=2002|ab=B|before=First Problem|num-a=2}} | |||
[[Category:Introductory Number Theory Problems]] | |||
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