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2002 AMC 10B Problems

Revision as of 18:37, 26 December 2008 by James4l (talk | contribs) (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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Problem 1

The ratio $2^{2001}\cdot{3^{2003}}\over6^{2002}$ is:

(A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 3/2

Solution

== Problem 2 ==For the nonzero numbers a, b, and c, define

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

Find (2,4,6).

(A) 1 (B) 2 (C) 4 (D) 6 (E) 24

Solution

Problem 3

Solution

Problem 4

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Problem 5

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Problem 6

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Problem 7

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Problem 8

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Problem 9

Solution

Problem 10

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Problem 11

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Problem 12

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Problem 13

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Problem 14

Solution

Problem 15

Solution

Problem 16

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Problem 17

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Problem 18

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Problem 19

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Problem 20

Solution

Problem 21

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Problem 22

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Problem 23

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Problem 24

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Problem 25

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See also