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

Latest revision as of 23:27, 29 November 2016

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:

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

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

$\fbox{D}$

1993 AHSME (Problems • Answer Key • Resources)
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

@@ Line 1: / Line 1: @@
-==Problem==
+== Problem ==
+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==
+<math>\text{(A) } -4\quad
+\text{(B) } -2\quad
+\text{(C) } 0\quad
+\text{(D) } 2\quad
+\text{(E) } 4</math>
-Plugging in to the equation, we have 1^-1 - -1^2 + 2^1 = 2.
+== 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>
+== See also ==
+{{AHSME box|year=1993|num-b=1|num-a=2}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}