Difference between revisions of "2009 AIME I Problems/Problem 6"
Kevinchang13 (talk | contribs) m (→Solution) |
Kevinchang13 (talk | contribs) m (→Solution) |
||
Line 6: | Line 6: | ||
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>. | ||
− | + | Because <math>{\lfloor x\rfloor}</math> must be an integer, we can do some simple case work: | |
For <math>{\lfloor x\rfloor}=0</math>, <math>N=1</math> no matter what <math>x</math> is | For <math>{\lfloor x\rfloor}=0</math>, <math>N=1</math> no matter what <math>x</math> is |
Revision as of 23:41, 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
.
Because must be an integer, we can do some simple 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.