Difference between revisions of "1998 AHSME Problems/Problem 5"
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− | == Problem | + | == Problem== |
If <math>2^{1998}-2^{1997}-2^{1996}+2^{1995} = k \cdot 2^{1995},</math> what is the value of <math>k</math>? | If <math>2^{1998}-2^{1997}-2^{1996}+2^{1995} = k \cdot 2^{1995},</math> what is the value of <math>k</math>? | ||
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==Solution 2== | ==Solution 2== | ||
− | Divide both sides of the original equation by <math>2^1995</math>, giving: | + | Divide both sides of the original equation by <math>2^{1995}</math>, giving: |
<math>2^3 - 2^2 - 2^1 + 1 = k</math> | <math>2^3 - 2^2 - 2^1 + 1 = k</math> | ||
Line 28: | Line 28: | ||
== See also == | == See also == | ||
{{AHSME box|year=1998|num-b=4|num-a=6}} | {{AHSME box|year=1998|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Latest revision as of 13:28, 5 July 2013
Contents
Problem
If what is the value of ?
Solution 1
Factor out :
Simplify:
By comparing the answer with the original equation, , and the answer is
Solution 2
Divide both sides of the original equation by , giving:
, and the answer is
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.