2002 AMC 10B Problem 1

Exponent rules only let you cancel powers of the same base. The top has 2s and 3s but the bottom has 6s, so first rewrite the bottom.

1. Since $6 = 2 \cdot 3$, multiplying $2002$ sixes together is the same as multiplying $2002$ twos and $2002$ threes:
$$6^{2002} = 2^{2002}\cdot 3^{2002}.$$
2. Now the fraction is
$$\frac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}}.$$
3. Cancel the twos. There are $2001$ on top and $2002$ on the bottom, so one $2$ is left on the bottom.
4. Cancel the threes. There are $2003$ on top and $2002$ on the bottom, so one $3$ is left on top.
5. What remains is $\dfrac{3}{2}$.

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

The huge exponents look scary, but only the differences between them matter once each base is the same prime. Whenever a problem mixes a base like $6$ with its prime pieces $2$ and $3$, break everything into primes first and then count what is left over.

Build sixes on top instead. Pair each of the $2001$ twos on top with a three: that uses $2001$ threes and makes $6^{2001}$, leaving $3^{2003-2001} = 3^2 = 9$. So the top is $9 \cdot 6^{2001}$. Dividing by $6^{2002}$ removes all those sixes and one more, giving $\frac{9}{6} = \frac{3}{2}$.