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

Revision as of 03:58, 3 September 2021

Problem

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

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

Solution 1

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

Solution 2

$\frac{2^{2001}\cdot3^{2003}}{6^{2002}}=\frac{2^{2001}\cdot 2\cdot 3^{2002}\cdot 3}{6^{2002}\cdot 2}=\frac{2^{2002} \cdot 3^{2002} \cdot 3}{6^{2002}\cdot 2}=\frac{6^{2002}\cdot 3}{6^{2002}\cdot 2}=\frac{3}{2}$ or $\mathrm{ (E) \ }$ ~by mathwiz0

Solution 3

$\frac{2^{2001}\cdot3^{2003}}{6^{2002}}$ $\frac{2^{2001}\cdot3^{2003}}{2^{2002}\cdot{3^{2002}}$ (Error compiling LaTeX. Unknown error_msg) $\frac{3}{2}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 15: / Line 15: @@
 == Solution 3 ==
-<math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}=\frac{2^{2001}\cdot3^{2003}}{2^{2002}\cdot{3^{2002}}=\frac{3}{2}</math>
+<math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}</math>
+<math>\frac{2^{2001}\cdot3^{2003}}{2^{2002}\cdot{3^{2002}}</math>
+<math>\frac{3}{2}</math>
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