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

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If <math>b=0</math>, the expression evaluates to <math>c-d\leq 2</math>. <br/>
 
If <math>b=0</math>, the expression evaluates to <math>c-d\leq 2</math>. <br/>
 
Case <math>d=0</math> remains.
 
Case <math>d=0</math> remains.
−
 
+
In that case, we want to maximize <math>c\cdot a^b</math> where <math>\{a,b,c\}=\{1,2,3\}</math>. Trying out the six possibilities we get that the best one is <math>(a,b,c)=(3,2,1)</math>, where <math>c\cdota^b=1\cdot 3^2=\boxed{\mathrm{(D)}\ 9}</math>.
−
In that case, we want to maximize <math>c\cdot a^b</math> where <math>\{a,b,c\}=\{1,2,3\}</math>. Trying out the six possibilities we get that the best one is <math>(a,b,c)=(3,2,1)</math>, where <math>c\cdot a^b = 1\cdot 3^2 = \boxed{9} \Longrightarrow \mathrm{(D)}</math>.
 
  
 
== See Also ==
 
== See Also ==

Revision as of 20:10, 22 July 2014

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 $c\cdota^b=1\cdot 3^2=\boxed{\mathrm{(D)}\ 9}$ (Error compiling LaTeX. Unknown error_msg).

See Also

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

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