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2004 AMC 10B Problem 5

2004 AMC 10B Problem 5AMC 10 Step by Step

Name: Date: October 28, 2026

Exponents, Logarithms & Radicals · difficulty 1 of 5 · about 2 min · position P1-10

In the expression $c\cdot a^b-d$ , the values of $a$ , $b$ , $c$ , and $d$ are $0$ , $1$ , $2$ , and $3$ , although not necessarily in that order. What is the maximum possible value of the result?

A) $5$
B) $6$
C) $8$
D) $9$
E) $10$

Answer

Key insight

Put the 0 where it does no damage: as the number being subtracted. Then try the six ways to place 1, 2, 3; one times 3 squared gives 9.

Solution

The $0$ does the most damage, so decide where it goes first.

If $a = 0$ or $c = 0$ , the product $c \cdot a^b$ is $0$ and the result is $0 - d$ , which is negative. Bad.
If $b = 0$ , then $a^0 = 1$ , and the result is $c - d$ . With $c$ and $d$ from $1, 2, 3$ , the best is $3 - 1 = 2$ . Small.
If $d = 0$ , nothing is subtracted, and the result is just $c \cdot a^b$ . This is clearly the best place for the $0$ .

Now place $1, 2, 3$ into $a, b, c$ . There are only six ways, so try them all:

$c$	$a$	$b$	$c \cdot a^b$
$1$	$3$	$2$	$1 \cdot 3^2 = 9$
$1$	$2$	$3$	$1 \cdot 2^3 = 8$
$2$	$3$	$1$	$2 \cdot 3^1 = 6$
$2$	$1$	$3$	$2 \cdot 1^3 = 2$
$3$	$2$	$1$	$3 \cdot 2^1 = 6$
$3$	$1$	$2$	$3 \cdot 1^2 = 3$

The biggest value is $9$ , with $a = 3$ , $b = 2$ , $c = 1$ , $d = 0$ .

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

Why this works

When there are only a handful of arrangements, a quick organized list beats clever guessing. A few facts help you see why the winner wins: a $0$ belongs where it is subtracted, a $1$ is wasted as a base since $1$ to any power is $1$ , and $3^2 = 9$ is a bit bigger than $2^3 = 8$ .

The trap

Stopping at 2 cubed = 8 (choice C) without checking 3 squared = 9.

Concise solution

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 $\boxed{\textbf{(D)}\ 9}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 1: Follow the Rules (§1.6, §1.7, §1.8)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.4)
Introduction to Algebra, Chapter 19: Exponents and Logarithms (§19.4)

Techniques

Organized listing / direct enumeration · Consider the largest/smallest element or boundary case