Difference between revisions of "2009 AIME I Problems/Problem 6"
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First, <math>x</math> must be less than <math>5</math>, since otherwise <math>x^{\lfloor x\rfloor}</math> would be at least <math>3125</math> which is greater than <math>1000</math>. | First, <math>x</math> must be less than <math>5</math>, since otherwise <math>x^{\lfloor x\rfloor}</math> would be at least <math>3125</math> which is greater than <math>1000</math>. | ||
| − | Now in order for <math>x^{\lfloor x\rfloor}</math> to be an integer, <math>x</math> must be an integral root of an integer, | + | Now in order for <math>x^{\lfloor x\rfloor}</math> to be an integer, <math>x</math> must be an integral root of an integer, so let do case work: |
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For <math>{\lfloor x\rfloor}=0</math>, N=<math>1</math> no matter what x is | For <math>{\lfloor x\rfloor}=0</math>, N=<math>1</math> no matter what x is | ||
Revision as of 14:01, 20 March 2009
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 in order for to be an integer, must be an integral root of an integer, so let do case work:
For 
, N=
no matter what x is
For 
, N can be anything between 
to 
excluding
This gives us 
N's
For 
, N can be anything between 
to 
excluding
This gives us 
N's
For 
, N can be anything between 
to 
excluding
This gives us 
N's
For 
, N can be anything between 
to 
excluding
This gives us 
N's
Since must be less than , we can stop here
Answer 
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 | ||
