2004 AMC 10B Problem 5

Decide where the $0$ goes. If $a = 0$ or $c = 0$, the product vanishes and the result is $-d$, negative. If $b = 0$, then $a^b = 1$ and the result is $c - d \le 3 - 1 = 2$. So the best use of $0$ is $d = 0$, which also removes the subtraction entirely.

Now distribute $1, 2, 3$ among $a, b, c$ to maximize $c \cdot a^b$. Test the six orders:

• $c = 1$: $3^2 = 9$ or $2^3 = 8$.
• $c = 2$: $2 \cdot 3^1 = 6$ or $2 \cdot 1^3 = 2$.
• $c = 3$: $3 \cdot 2^1 = 6$ or $3 \cdot 1^2 = 3$.

The largest is $1 \cdot 3^2 = 9$, from $(a, b, c, d) = (3, 2, 1, 0)$.

The answer is $\boxed{\textbf{(D)}\ 9}$.

With only four values and a tiny expression, organized enumeration is faster than clever reasoning. The structural facts that prune the search are worth remembering: a $0$ should sit where it does no harm (the subtracted term), a $1$ is wasted as a base, and among small powers $3^2 > 2^3$.

• Grabbing 2^3 = 8 as the largest power (choice (C)) without also testing 3^2 = 9.
• Putting the $1$ into $d$ "because it is small" and settling for $9 - 1 = 8$ or $6 - 1 = 5$, when $d = 0$ is available.