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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |