Difference between revisions of "2009 AIME I Problems/Problem 6"
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For <math>{\lfloor x\rfloor}=1</math>, <math>N</math> can be anything between <math>1^1</math> to <math>2^1</math> excluding <math>2^1</math> | For <math>{\lfloor x\rfloor}=1</math>, <math>N</math> can be anything between <math>1^1</math> to <math>2^1</math> excluding <math>2^1</math> | ||
− | Therefore, <math>N=1</math>. However, we got N=1 in case 1 so it got counted twice. | + | Therefore, <math>N=1</math>. However, we got <math>N=1</math> in case 1 so it got counted twice. |
For <math>{\lfloor x\rfloor}=2</math>, <math>N</math> can be anything between <math>2^2</math> to <math>3^2</math> excluding <math>3^2</math> | For <math>{\lfloor x\rfloor}=2</math>, <math>N</math> can be anything between <math>2^2</math> to <math>3^2</math> excluding <math>3^2</math> |
Revision as of 01:08, 27 November 2018
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 , 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.
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.