# 2002 AMC 10B Problems

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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

== 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