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

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{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #2]] and [[2004 AMC 10B Problems/Problem 5|2004 AMC 10B #5]]}}
 
{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #2]] and [[2004 AMC 10B Problems/Problem 5|2004 AMC 10B #5]]}}
 
== Problem 2 ==
 
== Problem 2 ==
In the expression <math>c\cdot a^b-d</math>, the values of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are 0, 1, 2, and 3, although not necessarily in that order. What is the maximum possible value of the result?
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In the expression <math>c\cdot a^b-d</math>, the values of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are <math>0</math>, <math>1</math>, <math>2</math>, and <math>3</math>, although not necessarily in that order. What is the maximum possible value of the result?
  
<math>(\mathrm {A}) 5\qquad (\mathrm {B}) 6 \qquad (\mathrm {C}) 8 \qquad (\mathrm {D}) 9 \qquad (\mathrm {E}) 10</math>
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<math>\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }8\qquad\mathrm{(D)\ }9\qquad\mathrm{(E)\ }10</math>
  
[[2004 AMC 12B Problems/Problem 2|Solution]]
 
 
== Solution ==
 
== Solution ==
  
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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.
 
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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 greatest is <math>(a,b,c)=(3,2,1)</math>, where <math>c\cdot a^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 ==
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{{AMC12 box|year=2004|ab=B|num-b=1|num-a=3}}
 
{{AMC12 box|year=2004|ab=B|num-b=1|num-a=3}}
 
{{AMC10 box|year=2004|ab=B|num-b=4|num-a=6}}
 
{{AMC10 box|year=2004|ab=B|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 13:04, 29 May 2021

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

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

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