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Difference between revisions of "1993 AHSME Problems/Problem 1"

(Created page with '==Problem== ==Solution== Plugging in to the equation, we have 1^-1 - -1^2 + 2^1 = 2.')
 
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==Problem==
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== Problem ==
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For integers <math>a,b,</math> and <math>c</math> define <math>\fbox{a,b,c}</math> to mean <math>a^b-b^c+c^a</math>. Then <math>\fbox{1,-1,2}</math> equals:
  
==Solution==
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<math>\text{(A) } -4\quad
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\text{(B) } -2\quad
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\text{(C) } 0\quad
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\text{(D) } 2\quad
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\text{(E) } 4</math>
  
Plugging in to the equation, we have 1^-1 - -1^2 + 2^1 = 2.
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== Solution ==
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Plug in the values for <math>a,b,c</math> and you get <math>1^{-1} - (-1)^2 + 2^1 \Rightarrow 1-1+2 \Rightarrow \fbox{2}</math>
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<math>\fbox{D}</math>
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== See also ==
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{{AHSME box|year=1993|num-b=1|num-a=2}} 
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 23:27, 29 November 2016

Problem

For integers $a,b,$ and $c$ define $\fbox{a,b,c}$ to mean $a^b-b^c+c^a$. Then $\fbox{1,-1,2}$ equals:

$\text{(A) } -4\quad \text{(B) } -2\quad \text{(C) } 0\quad \text{(D) } 2\quad \text{(E) } 4$

Solution

Plug in the values for $a,b,c$ and you get $1^{-1} - (-1)^2 + 2^1 \Rightarrow 1-1+2 \Rightarrow \fbox{2}$

$\fbox{D}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 2
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 26 27 28 29 30
All AHSME Problems and Solutions

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