Difference between revisions of "1993 AHSME Problems/Problem 1"
Epicfailiure (talk | contribs) (Created page with '==Problem== ==Solution== Plugging in to the equation, we have 1^-1 - -1^2 + 2^1 = 2.') |
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− | ==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: | ||
− | + | <math>\text{(A) } -4\quad | |
+ | \text{(B) } -2\quad | ||
+ | \text{(C) } 0\quad | ||
+ | \text{(D) } 2\quad | ||
+ | \text{(E) } 4</math> | ||
− | + | == 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}} |
Latest revision as of 23:27, 29 November 2016
Problem
For integers and define to mean . Then equals:
Solution
Plug in the values for and you get
See also
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 |
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