Difference between revisions of "2009 AIME I Problems/Problem 6"
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This gives us <math>5^4-4^4=369</math> <math>N</math>'s | This gives us <math>5^4-4^4=369</math> <math>N</math>'s | ||
− | Since <math>x</math> must be less than <math>5</math>, we can stop here and the | + | Since <math>x</math> must be less than <math>5</math>, we can stop here and the answer is <math>\boxed {4}</math> possible values for <math>N</math>. |
== See also == | == See also == | ||
{{AIME box|year=2009|n=I|num-b=5|num-a=7}} | {{AIME box|year=2009|n=I|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:40, 27 January 2015
Problem
How many positive integers less than are there such that the equation has a solution for ? (The notation denotes the greatest integer that is less than or equal to .)
Solution
First, must be less than , since otherwise would be at least which is greater than .
Now, must be an integer, so lets do case work:
For , no matter what is
For , can be anything between to excluding
This gives us 's
For , can be anything between to excluding
This gives us 's
For , can be anything between to excluding
This gives us 's
For , can be anything between to excluding
This gives us 's
Since must be less than , we can stop here and the answer is possible values for .
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.