2002 AMC 10B Problems/Problem 1: Difference between revisions
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<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 </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> | |||
==See Also== | ==See Also== | ||
Revision as of 12:40, 21 December 2019
Problem
The ratio 
is:
Solution 1
or 
Solution 2
or
See Also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
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| All AMC 10 Problems and Solutions | ||
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