Difference between revisions of "2009 AIME I Problems/Problem 6"
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First, <math>x</math> must be less than <math>5</math>, since otherwise <math>x^{\lfloor x\rfloor}</math> would be at least <math>3125</math> which is greater than <math>1000</math>. | First, <math>x</math> must be less than <math>5</math>, since otherwise <math>x^{\lfloor x\rfloor}</math> would be at least <math>3125</math> which is greater than <math>1000</math>. | ||
− | Now in order for <math>x^{\lfloor x\rfloor}</math> to be an integer, <math>x</math> must be an integral root of an integer, so | + | Now in order for <math>x^{\lfloor x\rfloor}</math> to be an integer, <math>x</math> must be an integral root of an integer, so lets do case work: |
− | For <math>{\lfloor x\rfloor}=0</math>, | + | For <math>{\lfloor x\rfloor}=0</math>, <math>N=1</math> no matter what <math>x</math> 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>, <math>N</math> 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> <math>N</math>'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>, <math>N</math> 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> <math>N</math>'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>, <math>N</math> 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> <math>N</math>'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>, <math>N</math> 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> <math>N</math>'s |
− | Since <math>x</math> must be less than <math>5</math>, we can stop here | + | Since <math>x</math> must be less than <math>5</math>, we can stop here and the answer answer is <math>1+5+37+369= \boxed {412}</math>. |
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== See also == | == See also == | ||
{{AIME box|year=2009|n=I|num-b=5|num-a=7}} | {{AIME box|year=2009|n=I|num-b=5|num-a=7}} |
Revision as of 14:05, 20 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 lets do case work:
For , no matter what is
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
For , can be anything between to excluding
This gives us 's
Since must be less than , we can stop here and the answer answer is .
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 |