Difference between revisions of "2004 AMC 12B Problems/Problem 2"

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Revision as of 18:51, 3 July 2013

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

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 $c\cdot a^b = 1\cdot 3^2 = \boxed{9} \Longrightarrow \mathrm{(D)}$.

See Also

2004 AMC 12B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png

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