2002 AMC 10B Problems/Problem 1: Difference between revisions
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<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> | <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 | ~by mathwiz0 | ||
== Solution 3 == | |||
<math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}=\frac{2^{2001}\cdot3^{2003}}{2^{2002}\cdot{3^{2002}}=\frac{3}{2}</math> | |||
==See Also== | ==See Also== | ||
Revision as of 03:57, 3 September 2021
Problem
The ratio 
is:
Solution 1
or 
Solution 2
or
~by mathwiz0
Solution 3
$\frac{2^{2001}\cdot3^{2003}}{6^{2002}}=\frac{2^{2001}\cdot3^{2003}}{2^{2002}\cdot{3^{2002}}=\frac{3}{2}$ (Error compiling LaTeX. Unknown error_msg)
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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