Difference between revisions of "2009 AIME I Problems/Problem 6"
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Answer <math>= 1+5+37+369= \boxed {412}</math> | Answer <math>= 1+5+37+369= \boxed {412}</math> | ||
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+ | == See also == | ||
+ | {{AIME box|year=2009|n=I|num-b=5|num-a=7}} |
Revision as of 22:45, 19 March 2009
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 |