2002 AMC 10B Problem 1 — Solution | LIVE by Po-Shen Loh

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2002 AMC 10B Problem 1

Attempt Problem 1 of the 2002 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

What is the value of the ratio

$\dfrac{2^{2001} \cdot 3^{2003}}{6^{2002}}?$

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{2}$

Answer: E

Concepts:exponent

Difficulty rating: 770

Solution:

Since $6^{2002} = 2^{2002}\cdot 3^{2002},$ the ratio becomes $\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}} = \dfrac{3}{2}.$

Thus, the correct answer is E.

Problem 2#2 →

Problem 1 in Other Years

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