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

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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:
 
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:
  
<math>\text{(A)} -4\quad
+
<math>\text{(A) } -4\quad
\text{(B)} -2\quad
+
\text{(B) } -2\quad
\text{(C)} 0\quad
+
\text{(C) } 0\quad
text{(D)} 2\quad
+
\text{(D) } 2\quad
\text{(E)} 4</math>
+
\text{(E) } 4</math>
  
 
== Solution ==
 
== Solution ==
 +
 +
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>
 +
 
<math>\fbox{D}</math>
 
<math>\fbox{D}</math>
  

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
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All AHSME Problems and Solutions

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