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>. | ||
− | Because <math>{\lfloor x\rfloor}</math> must be an integer, | + | Because <math>{\lfloor x\rfloor}</math> must be an integer, let’s do case work based on <math>{\lfloor x\rfloor}</math>: |
For <math>{\lfloor x\rfloor}=0</math>, <math>N=1</math> as long as <math>x \neq 0</math>. This gives us <math>1</math> value of <math>N</math>. | For <math>{\lfloor x\rfloor}=0</math>, <math>N=1</math> as long as <math>x \neq 0</math>. This gives us <math>1</math> value of <math>N</math>. |
Revision as of 20:10, 19 August 2020
Contents
Problem
How many positive integers less than are there such that the equation has a solution for ?
Solution
First, must be less than , since otherwise would be at least which is greater than .
Because must be an integer, let’s do case work based on :
For , as long as . This gives us value of .
For , can be anything between to excluding
Therefore, . However, we got in case 1 so it got counted twice.
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 .
Alternatively, one could find that the values which work are to get the same answer.
Solution 2
For a positive integer , we find the number of positive integers such that has a solution with . Then , and because , we have , and because is an integer, we get . The number of possible values of is equal to the number of integers between and inclusive, which is equal to the larger number minus the smaller number plus one or , and this is equal to . If , the value of exceeds , so we only need to consider . The requested number of values of is the same as the number of values of , which is .
Video Solution
Mostly the above solution explained on video: https://www.youtube.com/watch?v=2Xzjh6ae0MU&t=11s
~IceMatrix
Video Solution 2
~Shreyas S
Video Solution 3
Projective Solution: https://youtu.be/fUef_tVnM5M
~Shreyas S
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.