Difference between revisions of "2002 AMC 10B Problems/Problem 1"

Revision as of 10:22, 4 July 2013

The ratio $\frac{2^{2001}\cdot3^{2003}}{6^{2002}}$ is:

$\mathrm{(A) \ } 1/6\qquad \mathrm{(B) \ } 1/3\qquad \mathrm{(C) \ } 1/2\qquad \mathrm{(D) \ } 2/3\qquad \mathrm{(E) \ } 3/2$

$\frac{2^{2001}\cdot3^{2003}}{6^{2002}}=\frac{6^{2001}\cdot 3^2}{6^{2002}}=\frac{9}{6}=\frac{3}{2}$ or $\mathrm{ (E) \ }$

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Revision as of 12:54, 27 December 2008 (view source) Temperal (talk \| contribs) (bpx) ← Older edit		Revision as of 10:22, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
Line 13:		Line 13:

	[[Category:Introductory Number Theory Problems]]		[[Category:Introductory Number Theory Problems]]
		+	{{MAA Notice}}