2004 AMC 12B Problems/Problem 2

The following problem is from both the 2004 AMC 12B #2 and 2004 AMC 10B #5, so both problems redirect to this page.

Problem 2

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?

$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }8\qquad\mathrm{(D)\ }9\qquad\mathrm{(E)\ }10$

Solution

If $a=0$ or $c=0$ , the expression evaluates to $-d<0$ .
If $b=0$ , the expression evaluates to $c-d\leq 2$ .
Case $d=0$ remains.

In that case, we want to maximize $c\cdot a^b$ where $\{a,b,c\}=\{1,2,3\}$ . Trying out the six possibilities we get that the best one is $(a,b,c)=(3,2,1)$ , where $c\cdot a^b = 1\cdot 3^2 = \boxed{9} \Longrightarrow \mathrm{(D)}$ .

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2004 AMC 12B Problems/Problem 2

Problem 2

Solution

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