AMC10 2002 B

AMC10 2002 B · Q1

AMC10 2002 B · Q1. It mainly tests Fractions, Primes & prime factorization.

The ratio $\frac{2^{2001} \cdot 3^{2003}}{6^{2002}}$ is

比例 $\frac{2^{2001} \cdot 3^{2003}}{6^{2002}}$ 是

(A) $\frac{1}{6}$ $\frac{1}{6}$

(B) $\frac{1}{3}$ $\frac{1}{3}$

(D) $\frac{2}{3}$ $\frac{2}{3}$

(E) $\frac{3}{2}$ $\frac{3}{2}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

(E) We have $\dfrac{2^{2001}\cdot 3^{2003}}{6^{2002}}=\dfrac{2^{2001}\cdot 3^{2003}}{(2\cdot 3)^{2002}}=\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}}=\dfrac{3}{2}$

（E）我们有 $\dfrac{2^{2001}\cdot 3^{2003}}{6^{2002}}=\dfrac{2^{2001}\cdot 3^{2003}}{(2\cdot 3)^{2002}}=\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}}=\dfrac{3}{2}$

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