Difference between revisions of "2009 AIME I Problems/Problem 6"
(→Solution) |
|||
Line 10: | Line 10: | ||
So let do case work: | So let do case work: | ||
− | For <math>{\lfloor x\rfloor}=0</math> N=<math>1</math> no matter what x is | + | For <math>{\lfloor x\rfloor}=0</math>, N=<math>1</math> no matter what x is |
− | For <math>{\lfloor x\rfloor}=1</math> N 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>, N can be anything between <math>1^1</math> to <math>2^1</math> excluding <math>2^1</math> |
This gives us <math>2^1-1^1=1</math> N's | This gives us <math>2^1-1^1=1</math> N's | ||
− | For <math>{\lfloor x\rfloor}=2</math> N 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>, N can be anything between <math>2^2</math> to <math>3^2</math> excluding <math>3^2</math> |
This gives us <math>3^2-2^2=5</math> N's | This gives us <math>3^2-2^2=5</math> N's | ||
− | For <math>{\lfloor x\rfloor}=3</math> N can be anything between <math>3^3</math> to <math>4^3</math> excluding <math>4^3</math> | + | For <math>{\lfloor x\rfloor}=3</math>, N can be anything between <math>3^3</math> to <math>4^3</math> excluding <math>4^3</math> |
This gives us <math>4^3-3^3=37</math> N's | This gives us <math>4^3-3^3=37</math> N's | ||
− | For <math>{\lfloor x\rfloor}=4</math> N can be anything between <math>4^4</math> to <math>5^4</math> excluding <math>5^4</math> | + | For <math>{\lfloor x\rfloor}=4</math>, N can be anything between <math>4^4</math> to <math>5^4</math> excluding <math>5^4</math> |
This gives us <math>5^4-4^4=369</math> N's | This gives us <math>5^4-4^4=369</math> N's |
Revision as of 22:53, 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 |