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Difference between revisions of "2002 AMC 10B Problems"

(Problem 1: fixed LaTeX)
(LaTeXed some of the multiple choices)
Line 2: Line 2:
 
The ratio <math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}</math> is:
 
The ratio <math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}</math> is:
  
(A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 3/2
+
<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>
  
 
[[2002 AMC 10B Problems/Problem 1|Solution]]
 
[[2002 AMC 10B Problems/Problem 1|Solution]]
Line 12: Line 12:
  
 
Find <math>(2,4,6)</math>.
 
Find <math>(2,4,6)</math>.
 
+
<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 24 </math>
(A) 1 (B) 2 (C) 4 (D) 6 (E) 24
 
  
 
[[2002 AMC 10B Problems/Problem 2|Solution]]
 
[[2002 AMC 10B Problems/Problem 2|Solution]]

Revision as of 00:38, 30 December 2008

Problem 1

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

Problem 2

For the nonzero numbers a, b, and c, define

$(a,b,c)=\frac{abc}{a+b+c}$

Find $(2,4,6)$. $\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 24$

Solution

Problem 3

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

See also