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Difference between revisions of "2002 AMC 10B Problems"
(New page: ==Problem 1== The ratio <math>2^{2001}\cdot{3^{2003}}\over6^{2002}</math> is: (A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 3/2 Solution == Problem 2 ==For the...) |
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| Line 6: | Line 6: | ||
[[2002 AMC 10B Problems/Problem 1|Solution]] | [[2002 AMC 10B Problems/Problem 1|Solution]] | ||
| − | == Problem 2 ==For the nonzero numbers a, b, and c, define | + | == Problem 2 == |
| + | For the nonzero numbers a, b, and c, define | ||
(a,b,c)=<math>abc\over{a+b+c}</math> | (a,b,c)=<math>abc\over{a+b+c}</math> | ||
Revision as of 17:38, 26 December 2008
Contents
- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 See also
Problem 1
The ratio 
is:
(A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 3/2
Problem 2
For the nonzero numbers a, b, and c, define
(a,b,c)=
Find (2,4,6).
(A) 1 (B) 2 (C) 4 (D) 6 (E) 24
