2009 AIME I Problems/Problem 6
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 | |
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