Difference between revisions of "2002 AMC 10B Problems/Problem 1"
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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 1== | == Solution 1== | ||
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== Solution 3 == | == Solution 3 == | ||
| − | <math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}=\frac{2^{2001}\cdot3^{2003}}{2^{2002}\ | + | <math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}=\frac{2^{2001}\cdot3^{2003}}{2^{2002}\cdot3^{2002}}=\frac{3}{2}</math> |
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| + | ==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
Contents
Problem
The ratio 
is:
Solution 1
or 
Solution 2
or
~by mathwiz0
Solution 3
Video Solution by Daily Dose of Math
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
