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2002 AMC 10B Problems/Problem 1: Difference between revisions

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The ratio <math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}</math> is:
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\qquad \mathrm{(E) \ } 3/2 </math>
<math> \mathrm{(A) \ } 1/6\qquad \mathrm{(B) \ } 1/3\qquad \mathrm{(C) \ } 1/2\qquad \mathrm{(D) \ } 2/3\qquad \mathrm{(E) \ } 3/2\qquad</math>


== Solution ==
== Solution 1==


<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>
== Solution 2==
<math>\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}</math> or <math>\mathrm{ (E) \ }</math>
~by mathwiz0
== Solution 3 ==
<math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}=\frac{2^{2001}\cdot3^{2003}}{2^{2002}\cdot3^{2002}}=\frac{3}{2}</math>
==Video Solution by Daily Dose of Math==
https://youtu.be/d9tByrEEHuE
~Thesmartgreekmathdude


==See Also==
==See Also==

Latest revision as of 22:22, 24 October 2024

Problem

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

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

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

$\frac{2^{2001}\cdot3^{2003}}{6^{2002}}=\frac{2^{2001}\cdot3^{2003}}{2^{2002}\cdot3^{2002}}=\frac{3}{2}$

Video Solution by Daily Dose of Math

https://youtu.be/d9tByrEEHuE

~Thesmartgreekmathdude

See Also

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

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