Difference between revisions of "1989 AHSME Problems/Problem 1"
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| − | + | == Problem == | |
| − | (A) -7 | + | <math> (-1)^{5^{2}}+1^{2^{5}}= </math> |
| + | |||
| + | <math> \textrm{(A)}\ -7\qquad\textrm{(B)}\ -2\qquad\textrm{(C)}\ 0\qquad\textrm{(D)}\ 1\qquad\textrm{(E)}\ 57 </math> | ||
| + | |||
| + | == Solution == | ||
| + | |||
| + | <math>(-1)^{5^2} + 1^{2^5} = (-1)^{25}+1^{32}=-1+1=0</math> thus the answer is C. | ||
| + | == See also == | ||
| + | {{AHSME box|year=1989|before=First Question|num-a=2}} | ||
| + | |||
| + | [[Category: Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 06:37, 22 October 2014
Problem
Solution
thus the answer is C.
See also
| 1989 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
