2002 AMC 10B Problem 1

The denominator has a composite base, so rewrite it in primes: $6^{2002} = (2\cdot 3)^{2002} = 2^{2002}\cdot 3^{2002}$.

Now handle each prime on its own, subtracting the exponent below from the exponent above:
$$\frac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}} = 2^{2001-2002}\cdot 3^{2003-2002} = 2^{-1}\cdot 3^{1} = \frac{3}{2}.$$

The answer is $\boxed{\textbf{(E)}\ 3/2}$.

Exponent rules only combine powers of the same base, so a composite base like $6$ must be factored before anything cancels. Once every factor is a prime power, the enormous exponents are irrelevant; only the small differences $2001 - 2002$ and $2003 - 2002$ matter. Look for this whenever a problem pairs a composite base with primes.

Peel off matching powers: $2^{2001}\cdot 3^{2003} = (2\cdot 3)^{2001}\cdot 3^{2} = 9\cdot 6^{2001}$. Dividing by $6^{2002}$ leaves $\frac{9}{6} = \frac{3}{2}$.

• Getting the leftover exponents backwards (a spare 2 on top and a spare 3 below) and answering 2/3.
• Treating $2^{2001}\cdot 3^{2003}$ as $6^{4004}$ by adding exponents across different bases.